exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | 67173d9376a4aa8710851f8c9475e58f | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.Scanner;
public class PrincessesAndPrinces {
public static void main(String args[]) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t > 0) {
int n = in.nextInt();
boolean[] princes = new boolean[n];
int notSatisfiedDaughter = -1;
for (int i = 0; i < n; i++) {
int numChoices = in.nextInt();
boolean isMarried = false;
for (int j = 0; j < numChoices; j++) {
int choice = in.nextInt();
if (!isMarried) {
if (princes[choice - 1] == false) {
princes[choice - 1] = true;
isMarried = true;
}
}
}
if (!isMarried) {
notSatisfiedDaughter = i + 1;
}
}
if (notSatisfiedDaughter == -1) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
int i = 0;
while (i < n) {
if (princes[i] == false) {
System.out.println(notSatisfiedDaughter + " " + (i + 1));
break;
}
i++;
}
}
t--;
}
in.close();
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 699f308c3e67b56f64f592669975e3c2 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class code {
public static void main(String[] args) throws Exception {
Scanner sc= new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
HashSet<Integer> prince= new HashSet<Integer>();
for(int i=1; i<=n; i++) {
prince.add(i);
}
boolean[] princess=new boolean[n];
int c=0;
for(int j=0; j<n; j++) {
int k=sc.nextInt();
PriorityQueue<Integer> pq= new PriorityQueue<Integer>();
for(int i=0; i<k; i++)
pq.add(sc.nextInt());
for(int i=0; i<k; i++) {
int x= pq.poll();
if(prince.contains(x)) {
prince.remove(x);
princess[j]=true;
break;
}
}
if(!princess[j]) {
;
}
else
c++;
}
if(c==n) {
pw.println("OPTIMAL");
}
else {
boolean imp=false;
int i1=0;
int i2=0;
for(int j=princess.length-1; j>=0 && !imp; j--) {
if(!princess[j]) {
for(int i=1; i<=n && !imp; i++) {
if(prince.contains(i)) {
i1=j+1;
i2=i;
imp=true;
}
}
}
}
pw.println(imp? "IMPROVE": "OPTIMAL");
if(imp)
pw.println(i1+" "+i2);
}
}
pw.close();
}
static class Scanner {
StringTokenizer st;
BufferedReader br;
public Scanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
public Scanner(FileReader r) {
br = new BufferedReader(r);
}
public String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
public int nextInt() throws IOException {
return Integer.parseInt(next());
}
public long nextLong() throws IOException {
return Long.parseLong(next());
}
public String nextLine() throws IOException {
return br.readLine();
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public int[] nextIntArr(int n) throws IOException {
int[] arr = new int[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextInt();
return arr;
}
public boolean ready() throws IOException {
return br.ready();
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 715f2082aca29ca7bc05ad7f2b25ff14 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.Arrays;
public class BMarry {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine().trim());
while(t-->0) {
int n=Integer.parseInt(br.readLine().trim());
String str[];
boolean chek[]=new boolean[n+1];
int key=0;
for(int i=0;i<n;i++) {
str=br.readLine().trim().split(" ");
int k=Integer.parseInt(str[0]);
int arr[]=new int[k];
for(int j=0;j<k;j++) {
arr[j]=Integer.parseInt(str[j+1]);
}
Arrays.sort(arr);
int j=0;
for( j=0;j<k;j++) {
if(chek[arr[j]])continue;
else {
chek[arr[j]]=true;
break;
}
}
if(j==k) {
key=i+1;
}
}
int ans=0;
for(int i=1;i<=n;i++) {
if(!chek[i]) {
ans=i;
break;
}
}
if(key==0) {
System.out.println("OPTIMAL");
}else {
System.out.println("IMPROVE");
System.out.println(key+" "+ans);
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 2368cfe687627ae8fa23a85907e3085c | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class Princesses_and_Princes
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static int marry(TreeSet<Integer> s[],int n)
{
boolean used[]=new boolean[n+1];
for(int i=1;i<n+1;i++)
{
for(int x:s[i])
{
if(!used[x])
{
used[x]=true;
break;
}
}
}
int ans=0;
for(int i=1;i<n+1;i++)
{
if(used[i])
ans++;
}
return ans;
}
public static void main(String[] args)
{
FastReader sc=new FastReader();
PrintWriter pw=new PrintWriter(System.out);
int t=sc.nextInt();
m1:while(t-->0)
{
int n=sc.nextInt();
TreeSet<Integer> daughter[]=new TreeSet[n+1];
TreeSet<Integer> prince=new TreeSet<>();
for(int i=1;i<n+1;i++)
{
daughter[i]=new TreeSet<>();
int k=sc.nextInt();
prince.add(i);
for(int j=0;j<k;j++)
{
daughter[i].add(sc.nextInt());
}
}
boolean u[]=new boolean[n+1];
boolean m[]=new boolean[n+1];
m:for(int i=1;i<n+1;i++)
{
for(int x:daughter[i])
{
if(!u[x])
{
u[x]=true;
m[i]=true;
prince.remove(x);
continue m;
}
}
}
for(int i=1;i<n+1;i++)
{
if(!m[i] && !prince.isEmpty())
{
pw.println("IMPROVE");
pw.println(i+" "+prince.first());
continue m1;
}
}
pw.println("OPTIMAL");
}
pw.flush();
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 269c70ed58bffaf53989eee3b757da04 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class Two2 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
// System.out.println("n = "+n);
int[] princess = new int[n];
for (int i = 0; i < n; i++) {
princess[i] = -1;
}
Set<Integer> s = new HashSet<Integer>();
for (int i = 0; i < n; i++) {
int k = sc.nextInt();
// System.out.println("k = "+k);
boolean added = false;
for (int j = 0; j < k; j++) {
int p = sc.nextInt() - 1;
if (!added && princess[i] == -1 && !s.contains(p)) {
princess[i] = p;
s.add(p);
added = true;
}
}
}
/*
* for(int i=0;i<n;i++) { System.out.print(princess[i]+" "); }
*/
int princessNo = -1;
for (int i = 0; i < n; i++) {
if (princess[i] == -1) {
princessNo = i;
break;
}
}
int prince = -1;
for (int i = 0; i < n; i++) {
if (!s.contains(i)) {
//System.out.println("ftjf" + i);
prince = i;
break;
}
}
if (princessNo == -1 || prince == -1) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
System.out.println((princessNo + 1) + " " + (prince + 1));
}
}
sc.close();
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 310f5e890d48a5075a7ba5a74e582a3b | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
public class Solve {
FastScanner in;
PrintWriter out;
void solve() {
int T = in.nextInt();
while (0 < T--) {
int n = in.nextInt();
boolean[] a = new boolean[n];
boolean[] b = new boolean[n];
for (int i = 0; i < n; i++) {
int k = in.nextInt();
int[] g = in.nextIntArray(k);
for (int j = 0; j < k; j++) {
if ( ! b[g[j]-1]) {
a[i] = true;
b[g[j]-1] = true;
break;
}
}
}
// out.println(Arrays.toString(a));
// out.println(Arrays.toString(b));
int res1 = 0;
for (int i = 0; i < n; i++) {
if ( ! a[i]) {
res1 = i+1;
break;
}
}
int res2 = 0;
for (int i = 0; i < n; i++) {
if ( ! b[i]) {
res2 = i+1;
break;
}
}
if (res1 == 0 && res2 == 0) {
out.println("OPTIMAL");
} else {
out.println("IMPROVE");
out.println(res1 + " " + res2);
}
}
}
void run() {
try {
in = new FastScanner(new File("Solve.in"));
out = new PrintWriter(new File("Solve.out"));
solve();
out.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
void runIO() {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
solve();
out.close();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
boolean hasMoreTokens() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return false;
st = new StringTokenizer(s);
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextLong();
}
return a;
}
BigInteger nextBigInteger() {
return new BigInteger(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] nextDoubleArray(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextDouble();
}
return a;
}
}
public static void main(String[] args) {
new Solve().runIO();
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 4891cddcc08df876c480f0450c4cdb43 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
public class Solve {
FastScanner in;
PrintWriter out;
void solve() {
int T = in.nextInt();
while (0 < T--) {
int n = in.nextInt();
boolean[] a = new boolean[n];
boolean[] b = new boolean[n];
for (int i = 0; i < n; i++) {
int k = in.nextInt();
for (int j = 0; j < k; j++) {
int prince = in.nextInt();
if ( ! a[i] && ! b[prince-1]) {
a[i] = true;
b[prince-1] = true;
}
}
}
// out.println(Arrays.toString(a));
// out.println(Arrays.toString(b));
int res1 = -1;
int res2 = -1;
for (int i = 0; i < n; i++) {
if (res1 == -1 && ! a[i]) {
res1 = i+1;
}
if (res2 == -1 && ! b[i]) {
res2 = i+1;
}
}
if (res1 != -1 && res2 != -1) {
out.println("IMPROVE");
out.println(res1 + " " + res2);
} else {
out.println("OPTIMAL");
}
}
}
void run() {
try {
in = new FastScanner(new File("Solve.in"));
out = new PrintWriter(new File("Solve.out"));
solve();
out.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
void runIO() {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
solve();
out.close();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
boolean hasMoreTokens() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return false;
st = new StringTokenizer(s);
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextLong();
}
return a;
}
BigInteger nextBigInteger() {
return new BigInteger(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] nextDoubleArray(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextDouble();
}
return a;
}
}
public static void main(String[] args) {
new Solve().runIO();
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | c45a33090749541cf9b3351551246d31 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
public class Solve {
FastScanner in;
PrintWriter out;
void solve() {
int T = in.nextInt();
while (0 < T--) {
int n = in.nextInt();
boolean[] a = new boolean[n+1];
boolean[] b = new boolean[n+1];
for (int i = 1; i <= n; i++) {
int k = in.nextInt();
while (0 < k--) {
int g = in.nextInt();
if ( ! a[i] && ! b[g]) {
a[i] = true;
b[g] = true;
}
}
}
int res1 = Integer.MAX_VALUE;
int res2 = Integer.MAX_VALUE;
for (int i = 1; i <= n; i++) {
if (! a[i]) {
res1 = Math.min(res1, i);
}
if (! b[i]) {
res2 = Math.min(res2, i);
}
}
if (res1 != Integer.MAX_VALUE &&
res2 != Integer.MAX_VALUE) {
out.println("IMPROVE");
out.println(res1 + " " + res2);
} else {
out.println("OPTIMAL");
}
}
}
void run() {
try {
in = new FastScanner(new File("Solve.in"));
out = new PrintWriter(new File("Solve.out"));
solve();
out.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
void runIO() {
in = new FastScanner(System.in);
out = new PrintWriter(System.out);
solve();
out.close();
}
class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(File f) {
try {
br = new BufferedReader(new FileReader(f));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public FastScanner(InputStream f) {
br = new BufferedReader(new InputStreamReader(f));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return null;
st = new StringTokenizer(s);
}
return st.nextToken();
}
boolean hasMoreTokens() {
while (st == null || !st.hasMoreTokens()) {
String s = null;
try {
s = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (s == null)
return false;
st = new StringTokenizer(s);
}
return true;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextLong();
}
return a;
}
BigInteger nextBigInteger() {
return new BigInteger(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
double[] nextDoubleArray(int n) {
double[] a = new double[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextDouble();
}
return a;
}
}
public static void main(String[] args) {
new Solve().runIO();
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 9618ae63a3af2eefdefe7bcaa6ed3c13 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;
public class PrincessesAndPrinces {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
int [][] arr = new int[n][];
for (int i=0;i<n;i++) {
int k = s.nextInt();
arr[i] = new int[k];
for (int j=0;j<k;j++)
arr[i][j] = s.nextInt();
}
Set<Integer> set1 = new HashSet<>();
Set<Integer> set2 = new HashSet<>();
for (int i=1;i<=n;i++) {
set1.add(i);
set2.add(i);
}
int count = 0;
int a = -1;
for (int i=0;i<n;i++) {
boolean flag = true;
for (int j=0;j<arr[i].length;j++) {
int x = arr[i][j];
if (set2.contains(x)) {
set2.remove(x);
set1.remove(i+1);
flag = false;
count++;
break;
}
}
if (flag && a == -1)
a = i+1;
}
if (count == n)
System.out.println("OPTIMAL");
else {
for (int i=1;i<=n;i++) {
if (set2.contains(i)) {
System.out.println("IMPROVE");
System.out.println(a + " " + i);
break;
}
}
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | e5c874bc0a6e84acf9d1b83aeabefdc6 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
StringBuffer out = new StringBuffer();
int t=in.nextInt();
HashSet<Integer> set = new HashSet<>();
label2:
while(t--!=0) {
int n = in.nextInt();
ArrayList<Integer> list[] = new ArrayList[n+1];
for(int i=1; i<=n; i++) {
list[i] = new ArrayList<>();
for(int j=in.nextInt(); j>0; j--)
list[i].add(in.nextInt());
set.add(i);
}
boolean flag[] = new boolean[n+1];
label:
for(int i=1; i<=n; i++)
for(int item: list[i])
if(set.contains(item)) {
set.remove(item);
flag[i]=true;
continue label;
}
for(int i=1; i<=n; i++)
if(!flag[i]) {
out.append("IMPROVE\n");
int j = -1;
for(int item: set) {
j=item;
break;
}
out.append(i+" "+j+"\n");
continue label2;
}
out.append("OPTIMAL\n");
}
System.out.println(out);
}
public static int modInverse(int a, int m)
{
int m0 = m;
int y = 0, x = 1;
if (m == 1)
return 0;
while (a > 1)
{
int q = a / m;
int t = m;
m = a % m;
a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | e45cd08fa1504b2448fc4e3cfd2f5dab | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.util.*;
import java.io.*;
public class PrincessesAndPrinces {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
// Scanner scan = new Scanner(System.in);
FastIO scan = new FastIO(System.in);
int test = scan.nextInt();
while (test-- != 0) {
int n = scan.nextInt();
boolean f = true;
int taken = 0, m1 = 0;
HashSet<Integer> m = new HashSet<>();
m.add(Integer.MAX_VALUE);
//nloop
for (int i = 1; i <= n; i++) {
int k = scan.nextInt();
int min1 = Integer.MAX_VALUE;
boolean flag = true;
for (int j = 0; j < k; j++) {
int l = scan.nextInt();
if (!m.contains(l) && min1 > l) {
min1 = l;
flag = false;
}
}
if (!m.contains(min1)) {
m.add(min1);
} else {
taken = i;
}
}
for (int l = 1; l <= n; l++) {
if (m.contains(l)) {
} else {
f = false;
m1 = l;
}
}
if (!f == true) {
System.out.println("IMPROVE");
System.out.println((taken) + " " + m1);
} else {
System.out.println("OPTIMAL");
}
}
}
}
class FastIO {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new RuntimeException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new RuntimeException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public String readString() {
final StringBuilder stringBuilder = new StringBuilder();
int c = read();
while (isSpaceChar(c)) {
c = read();
}
do {
stringBuilder.append(c);
c = read();
} while (!isSpaceChar(c));
return stringBuilder.toString();
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | b5746cef40b60a4da7f3f6de518e92ea | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class Friend {
public static void main(String[] args) throws Exception{
// TODO Auto-generated method stub
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while (t>0)
{
int n=Integer.parseInt(br.readLine());
ArrayList<TreeSet<Integer>> arr=new ArrayList<>();
for (int i=0;i<n;i++)
{
TreeSet<Integer> temp=new TreeSet<>();
String inp[]=br.readLine().split(" ");
for (int j=1;j<inp.length;j++)
{
int x=Integer.parseInt(inp[j])-1;
temp.add(x);
}
arr.add(temp);
}
boolean prince[]=new boolean[n];
boolean daughter[]=new boolean[n];
int married=0;
for (int i=0;i<n;i++)
{
TreeSet<Integer> tree=arr.get(i);
for (int j : tree)
{
if (prince[j]==false)
{
prince[j]=true;
daughter[i]=true;
married++;
break;
}
}
}
if (married==n)
{
System.out.println("OPTIMAL");
t--;
continue;
}
int unmarriedprince=-1;
int undaughter=-1;
for (int i=0;i<n;i++)
{
if (prince[i]==false)
{
unmarriedprince=i;
break;
}
}
for (int i=0;i<n;i++)
{
if (daughter[i]==false)
{
if (arr.get(i).contains(unmarriedprince)==false)
{
undaughter=i;
break;
}
}
}
if (undaughter!=-1)
{
undaughter++;
unmarriedprince++;
System.out.println("IMPROVE");
System.out.println(undaughter+" "+unmarriedprince);
}
else
System.out.println("OPTIMAL");
t--;
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 65673ba7688ed4c11e372d597eb62923 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.stream.IntStream;
public class _E84_B implements Runnable{
public static void main(String[] args) {
new Thread(null, new _E84_B(),"Main",1<<27).start();
}
@Override
public void run() {
FastReader fd = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int t = fd.nextInt();
for(int te = 0; te < t; te++){
int n = fd.nextInt();
int remainingPrincess = 0;
boolean[] princes = new boolean[n+1];
for (int i = 0; i < n; i++) {
int listSize = fd.nextInt();
boolean isFixed = false;
for (int j = 0; j < listSize; j++) {
int min = fd.nextInt();
if (!isFixed) {
if (!princes[min]) { princes[min] = isFixed = true; }
}
}
if (!isFixed) { remainingPrincess = i+1; }
}
if (remainingPrincess == 0) {
out.println("OPTIMAL");
}
else {
out.println("IMPROVE");
for (int i = 1; i <= n; i++) {
if (!princes[i]) {
out.println(remainingPrincess + " " + i);
break;
}
}
}
}
out.close();
}
//Helper functions
static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
static int lcm(int a, int b) {
return (a*b)/gcd(a, b);
}
static boolean checkDistinct(int next){
String process = String.valueOf(next);
for(int i = 0;i < process.length()-1; i++){
String character = String.valueOf(process.charAt(i));
if(process.substring(i+1).contains(character)){ return false; }
}
return true;
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | a0e2400d18f62c6e460ee3b2dc6ce4b5 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.PrintWriter;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Random;
import java.util.Scanner;
public class TaskA {
static Scanner in = new Scanner(System.in);
static PrintWriter w = new PrintWriter(System.out);
public static void main(String[] args) {
int t = Integer.parseInt(in.nextLine());
outer: while (t-- > 0) {
int n = in.nextInt();
boolean[] arr = new boolean[n+1];
boolean[] mis = new boolean[n+1];
for(int i = 1; i<=n; i++) {
int temp = in.nextInt();
for(int j = 0; j<temp; j++) {
int prince = in.nextInt();
if(arr[prince]==false && mis[i]==false) {
arr[prince] = true;
mis[i] = true;
}
}
}
int ans1 = -1, ans2 = -1;
for(int i = 1; i<=n; i++) {
if(arr[i]==false && ans1 ==-1)
ans1 = i;
if(mis[i]==false && ans2 == -1)
ans2 =i;
}
w.println(ans1 == -1 && ans2 == -1 ? "OPTIMAL" : "IMPROVE\n" + ans2 + " " + ans1);
}
w.flush();
w.close();
}
private static void shuffleArray(int[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
int tmp = arr[i];
int randomPos = rnd.nextInt(n);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
private static void shuffleArray(long[] arr) {
int n = arr.length;
Random rnd = new Random();
for (int i = 0; i < n; ++i) {
long tmp = arr[i];
int randomPos = i + rnd.nextInt(n - i);
arr[i] = arr[randomPos];
arr[randomPos] = tmp;
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 2049ace15743e1cc62865b5ba503c441 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
//package graphtheory;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Set;
import java.util.TreeSet;
public class HackerEarth{
static int mod=1000000007;
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static int power(int x, int y, int p)
{
int res = 1;
x = x % p;
while (y > 0)
{
if((y & 1)==1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
// Returns n^(-1) mod p
static int modInverse(int n, int p)
{
return power(n, p-2, p);
}
public static void solve() throws IOException{
Reader sc = new Reader();
OutputStream outputStream = System.out;
PrintWriter out = new PrintWriter(outputStream);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
ArrayList<Integer> list=new ArrayList<>();
TreeSet<Integer> set=new TreeSet<>();
LinkedList<Integer> arr[]=new LinkedList[n+1];
for(int i=0;i<=n;i++){
arr[i]=new LinkedList<>();
}
for(int i=1;i<=n;i++){
int k=sc.nextInt();
// System.out.println(k+" i "+i);
for(int j=1;j<=k;j++){
arr[i].add(sc.nextInt());
}
//System.out.println(arr[i]+" "+i);
boolean take=false;
Iterator<Integer> itr=arr[i].iterator();
while(itr.hasNext()){
int next=(int)itr.next();
if(!set.contains(next)){
set.add(next);
take=true;
break;
}
}
//System.out.println(set+" "+i+" "+n);
if(!take){
list.add(i);
}
}
int q=0;
if(list.size()!=0)
q=list.get(0);
int p=0;
for(int i=1;i<=n;i++){
if(!set.contains(i)){
p=i;
break;
}
}
// System.out.println(list);
if(set.size()==n)
out.println("OPTIMAL");
else{
out.println("IMPROVE");
out.println(q+" "+p);
}
}
out.close();
}
public static void main(String args[] ) throws Exception {
new Thread(null, new Runnable() {
@Override
public void run() {
try {
solve();
} catch (Exception e) {
}
}
}, "1", 1<<26).start();
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[1000]; // line length
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public String nextString() throws IOException{
byte c = read();
while(Character.isWhitespace(c)){
c = read();
}
StringBuilder builder = new StringBuilder();
builder.append((char)c);
c = read();
while(!Character.isWhitespace(c)){
builder.append((char)c);
c = read();
}
return builder.toString();
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | ec08b01a620e2c9a65e471cd409599b7 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.io.*;
public class main
{
public static void main(String args[])throws java.lang.Exception
{
BufferedReader inp = new BufferedReader (new InputStreamReader(System.in));
int n=Integer.parseInt(inp.readLine());
while(n-->0)
{
int a=Integer.parseInt(inp.readLine());
boolean ans[]=new boolean[a+1];
boolean daughter[]=new boolean[a+1];
int da=0;
for(int i=0;i<a;i++)
{
String w[]=inp.readLine().split(" ");
int flag=0;
for(int k=1;k<w.length;k++)
{
if(!ans[Integer.parseInt(w[k])]){flag=1;ans[Integer.parseInt(w[k])]=true;daughter[i+1]=true;break;}
}
if(flag==0 && da==0){da=i+1;}
}
if(da!=0){
int aa=0;
for(int i=1;i<=a;i++){if(!ans[i]){aa=i;break;}}
System.out.println("IMPROVE");
System.out.println(da+" "+aa);
}
else{System.out.println("OPTIMAL");}
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 1844b07d8ffd8bf197c71bedc47cde03 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
sc.nextLine();
for(int x = 0; x < t; x++) {
int n = sc.nextInt();
sc.nextLine();
ArrayList<Integer>[] sets = new ArrayList[n+1];
for(int i = 1; i <= n; i++) {
sets[i] = new ArrayList<Integer>();
int amt = sc.nextInt();
for(int j = 0; j < amt; j++) {
sets[i].add(sc.nextInt());
}
sc.nextLine();
}
boolean[] taken = new boolean[n+1];
int change = 0;
boolean optimal = true;
for(int i = 1; i <= n; i++) {
boolean bad = true;
for(int a: sets[i]) {
if(!taken[a]) {
bad = false;
taken[a] = true;
break;
}
}
if(bad && optimal) {
change = i;
optimal = false;
}
}
int to = 0;
if(!optimal) {
for(int i = 1; i <= n; i++) {
if(!taken[i]) {
to = i;
break;
}
}
System.out.println("IMPROVE");
System.out.println(change + " " + to);
} else {
System.out.println("OPTIMAL");
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 29188d38d6257b8e5412c20d2bd62c64 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
public class Main
{
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int j=0;j<t;j++)
{
int n=sc.nextInt();
ArrayList<ArrayList<Integer>> a=new ArrayList<ArrayList<Integer>>(n);
for(int w=0;w<n;w++){
int k=sc.nextInt();
ArrayList<Integer> p=new ArrayList<Integer>();
for(int i=0;i<k;i++)
{
int z=sc.nextInt();
p.add(z);
}
Collections.sort(p);
a.add(p);
}
/*for(int i=0;i<a.size();i++)
{
for(int k=0;k<a.get(i).size();k++)
System.out.print(a.get(i).get(k)+" ");
System.out.println();
}
System.out.println();*/
boolean girl[]=new boolean[n+1];
for(int i=0;i<n+1;i++)
girl[i]=false;
boolean prince[]=new boolean[n+1];
for(int i=0;i<n+1;i++)
prince[i]=false;
for(int i=0;i<n;i++)
{
for(int k=0;k<a.get(i).size();k++)
{
if(prince[a.get(i).get(k)]==false){
prince[a.get(i).get(k)]=true;
girl[i+1]=true;
break;
}
}
}
/* for(int i=1;i<n+1;i++)
System.out.print(girl[i]+" ");
System.out.println();
boolean prince[]=new boolean[n+1];
for(int i=1;i<n+1;i++)
System.out.print(prince[i]+" ");
System.out.println();*/
int x=-1;
for(int i=1;i<n+1;i++)
{
if(girl[i]==false){
x=i;
break;
}
}
int y=-1;
for(int i=1;i<n+1;i++)
{
if(prince[i]==false){
y=i;
break;
}
}
if(x==-1)
System.out.println("OPTIMAL");
else{
System.out.println("IMPROVE");
System.out.println(x+" "+y);
}
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 0ea380727a57ecf0732600af4fc1a8eb | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
solve(in, out);
out.close();
}
public static void solve(InputReader in, PrintWriter out) {
int t = in.nextInt();
for (int i = 0; i < t; ++i) {
int n = in.nextInt();
Set<Integer> set = new HashSet<>(n);
int nonMarriedDaughter = -1;
for (int j = 0; j < n; ++j) {
int k = in.nextInt();
boolean married = false;
for (int jj = 0; jj < k; ++jj) {
int current = in.nextInt();
if (!set.contains(current)) {
married = true;
set.add(current);
in.skip();
break;
}
}
if (!married && nonMarriedDaughter == -1) {
nonMarriedDaughter = j;
}
}
if (nonMarriedDaughter == -1) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
for (int j = 1; j <= n; ++j) {
if (!set.contains(j)) {
System.out.print(nonMarriedDaughter + 1);
System.out.print(" ");
System.out.println(j);
break;
}
}
}
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public void skip() {
tokenizer = null;
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 7f62449a504282028bf2bc4aaeefdb40 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
import java.lang.Math;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader b1 = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(b1.readLine());
t: while(t-->0) {
int n = Integer.parseInt(b1.readLine());
boolean [] taken = new boolean[n+1];
int left = 0;
int right = 0;
boolean condition = false;
boolean mark = false;
int daughter = 0;
for(int i = 0; i<n; i++) {
int min = Integer.MAX_VALUE;
StringTokenizer s1 = new StringTokenizer(b1.readLine());
int count = 0;
while(s1.hasMoreTokens()) {
int val = Integer.parseInt(s1.nextToken());
if(!taken[val] && count > 0) {
min = Math.min(min, val);
}
count++;
//System.out.println("val: " + val);
}
//System.out.println("n: " + n);
if(min != Integer.MAX_VALUE && !taken[min]) {
taken[min] = true;
}
else {
mark = true;
daughter = i;
}
}
if(mark) {
for(int j = 1; j<taken.length; j++) {
if(taken[j] == false) {
condition = true;
left = daughter+1;
right = j;
}
}
}
if(condition) {
System.out.println("IMPROVE");
System.out.println((left) + " " + (right));
continue t;
}
System.out.println("OPTIMAL");
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 10958096a2e45dce4bbe4e5b97a818fe | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.util.Set;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskB solver = new TaskB();
solver.solve(1, in, out);
out.close();
}
static class TaskB {
public void solve(int testNumber, FastReader in, PrintWriter out) {
int t = in.nextInt();
while (t > 0) {
int n = in.nextInt();
Set<Integer> kingdoms = new HashSet<>();
for (int i = 1; i <= n; i++) {
kingdoms.add(i);
}
int princess = -1;
for (int i = 1; i <= n; i++) {
int x = in.nextInt();
boolean married = false;
for (int j = 0; j < x; j++) {
int prince = in.nextInt();
if (!married && kingdoms.contains(prince)) {
kingdoms.remove(prince);
married = true;
}
}
if (!married && princess == -1) {
princess = i;
}
}
if (princess != -1) {
out.println("IMPROVE");
for (Integer prince : kingdoms) {
out.println(princess + " " + prince);
break;
}
} else {
out.println("OPTIMAL");
}
t--;
}
}
}
static class FastReader {
BufferedReader in = null;
StringTokenizer st;
public FastReader() {
}
public FastReader(InputStream is) {
in = new BufferedReader(new InputStreamReader(is));
}
public String next() {
try {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
} catch (Exception e) {
return null;
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 246d67b35288408b7659d0726b4672ce | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.util.Map;
import java.util.Set;
import java.util.HashMap;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskB solver = new TaskB();
solver.solve(1, in, out);
out.close();
}
static class TaskB {
public void solve(int testNumber, FastReader in, PrintWriter out) {
int t = in.nextInt();
while (t > 0) {
int n = in.nextInt();
Set<Integer> kingdoms = new HashSet<>();
Map<Integer, Set<Integer>> unmarriedDaughters = new HashMap<>();
for (int i = 1; i <= n; i++) {
kingdoms.add(i);
}
for (int i = 1; i <= n; i++) {
int x = in.nextInt();
Set<Integer> choices = new HashSet<>();
boolean married = false;
for (int j = 0; j < x; j++) {
int king = in.nextInt();
if (!married && kingdoms.contains(king)) {
married = true;
kingdoms.remove(king);
}
choices.add(king);
}
if (!married) {
unmarriedDaughters.put(i, choices);
}
}
boolean done = false;
if (!unmarriedDaughters.isEmpty() && !kingdoms.isEmpty()) {
for (Integer i : unmarriedDaughters.keySet()) {
for (Integer king : kingdoms) {
if (!unmarriedDaughters.get(i).contains(king)) {
out.println("IMPROVE");
out.println(i + " " + king);
done = true;
break;
}
}
if (done) {
break;
}
}
}
if (!done) {
out.println("OPTIMAL");
}
t--;
}
}
}
static class FastReader {
BufferedReader in = null;
StringTokenizer st;
public FastReader() {
}
public FastReader(InputStream is) {
in = new BufferedReader(new InputStreamReader(is));
}
public String next() {
try {
while (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
} catch (Exception e) {
return null;
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | e1779ce9b1052c9efaf9cbfc586eb94a | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.lang.*;
// https://codeforces.com/problemset/problem/1327/B
public class B1327 {
static Scanner sc;
public static void solve() {
int n = sc.nextInt();
boolean[] d = new boolean[n+1];
boolean[] p = new boolean[n+1];
int missing = -1;
for(int i = 1; i <= n; i++) {
int x = sc.nextInt();
boolean done = false;
while(x-- > 0) {
int cp = sc.nextInt();
if(d[i] == true || p[cp] == true) continue;
d[i] = true;
p[cp] = true;
done = true;
}
if(! done) missing = i;
}
if(missing == -1) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
System.out.print(missing + " ");
for(int i = 1; i <= n; i++)
if(!p[i]) {
System.out.println(i);
return;
}
}
}
public static void main(String[] args) {
sc = new Scanner(System.in);
int n = sc.nextInt();
while(n-- > 0) solve();
System.out.println("");
sc.close();
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | a960d8080658416daaa7c405b274d512 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.util.*;
public class A_B_and_team_trainin{
public static void main(String[] args){
Scanner dek = new Scanner(System.in);
int t = dek.nextInt();
StringBuilder res = new StringBuilder();
while (t-- > 0) {
for(int ll=0;ll<2;ll++){
int aa=2+2*2;
}
int mn = dek.nextInt();
HashSet<Integer> mset = new HashSet<>();
int mo = -1;
for (int i = 1; i <= mn; i++) {
int k = dek.nextInt();
boolean b = false;
for (int j = 0; j < k; j++) {
int mmtemp = dek.nextInt();
if (!mset.contains(mmtemp) && !b) {
b = true;
mset.add(mmtemp);
}
}
if (!b && mo == -1) {
mo = i;
}
for(int ll=0;ll<2;ll++){
int aa=2+2*2;
}
}
// System.out.println(mset);
if (mset.size() == mn) {
res.append("OPTIMAL\n");
} else {
int i = 1;
for (; i <= mn; i++) {
if (mset.contains(i) == false)
break;
}
res.append("IMPROVE\n");
res.append(mo + " " + i + "\n");
}
for(int ll=0;ll<2;ll++){
int aa=2+2*2;
}
}
System.out.println(res);
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | c393770400d8703e8a58acd60f8a0d7f | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.util.ArrayList;
public class Main{
public static void main(String[] args){
Scanner param = new Scanner(System.in);
int end=param.nextInt();
while(end-->0){
int n=param.nextInt();
int arr[]=new int[n];
int p=-1;
boolean b=false;
HashSet<Integer>h=new HashSet<>();
int pp=-1;
ArrayList<Integer>l3=new ArrayList<>();
for(int i=0;i<n;i++){
int x=param.nextInt();
ArrayList<Integer>l1=new ArrayList<>();
b=false;
for(int t=0;t<x;t++){
int y=param.nextInt();
l1.add(y);
}
int j=0;
while(j<l1.size()){
if(!h.contains(l1.get(j))){
h.add(l1.get(j));
b=true;
break;
}
j++;
}
if(b==false){
l3.add(i);
}
}
for(int i=1;i<=n;i++){
if(!h.contains(i)){
pp=i;
break;
}
}
if(l3.size()==0&&pp==-1){
System.out.println("OPTIMAL");
}
else{
System.out.println("IMPROVE");
System.out.println((l3.get(0)+1)+" "+pp);
}
}
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 9f4def1ec9de97e6e6d0ad48d2f4c933 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.StringTokenizer;
public class B_Princesses {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int testCases = sc.nextInt();
while(testCases-- > 0) {
int pMax = sc.nextInt();
HashSet<Integer> unmarriedP = new HashSet<>();
HashSet<Integer> unmarriedK = new HashSet<>();
for(int i = 1; i <= pMax; i++) {
unmarriedK.add(i);
}
for(int i = 1; i <= pMax; i++) {
int minK = Integer.MAX_VALUE;
int edgeCount = sc.nextInt();
for(int j = 0; j < edgeCount; j++) {
int edge = sc.nextInt();
if(unmarriedK.contains(edge))
minK = Math.min(minK, edge);
}
if(minK != Integer.MAX_VALUE) {
unmarriedK.remove(minK);
}else {
unmarriedP.add(i);
}
}
if(unmarriedP.isEmpty()) {
out.println("OPTIMAL");
}else {
out.println("IMPROVE");
out.println(unmarriedP.iterator().next() + " " + unmarriedK.iterator().next());
}
}
out.close();
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine(){
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | e525fb2bc6a343da4692444a7574f13a | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | //package com.company;
import java.io.*;
import java.util.*;
import java.lang.*;
public class Main{
public static class FastReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public FastReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars==-1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
}
catch (IOException e) {
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
public int nextInt() {
int c = read();
while(isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if(c<'0'||c>'9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c))
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public char nextChar() {
int c = read();
while (isSpaceChar(c))
c = read();
return (char)c;
}
public String next() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
static long gcd(long a, long b){
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a*b)/gcd(a, b);
}
public static void sortbyColumn(int arr[][], int col)
{
Arrays.sort(arr, new Comparator<int[]>() {
@Override
public int compare(final int[] entry1,
final int[] entry2) {
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
});
}
static LinkedList<Integer> li[]=new LinkedList[100001];
static int ans1=0;
static int visit[]=new int[100001];
static void dfs(int x){
visit[x]=1;
ans1++;
for(int i=0;i<li[x].size();i++){
if(visit[li[x].get(i)]==0){
dfs(li[x].get(i));
}
}
}
public static void main(String args[]){
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
FastReader in = new FastReader(inputStream);
PrintWriter w = new PrintWriter(outputStream);
int t,i,j,k;
t=in.nextInt();
for(i=0;i<t;i++){
int n,m;
n=in.nextInt();
//m=in.nextInt();
TreeSet<Integer> tr=new TreeSet<>();
for(j=0;j<n;j++){
tr.add(j+1);
}
int flag1=0,flag2=0,x=0,y=0;
int flag[]=new int[n];
for(j=0;j<n;j++){
m=in.nextInt();
flag2=0;
for(k=0;k<m;k++){
int z=in.nextInt();
if(flag2!=1){
if(tr.contains(z)){
tr.remove(z);
//flag1=1;
flag[j]=1;
flag2=1;
//w.println("here at "+k+" "+j);
}
}
}
}
for(j=0;j<n;j++){
if(flag[j]==0){
x=j+1;
if(!tr.isEmpty())
y=tr.first();
flag1=1;
break;
}
}
// if(flag1!=1){
// if(flag2==0){
// x=j+1;
// y=tr.first();
// flag1=1;
// }
// }
// }
if(flag1==1){
w.println("IMPROVE");
w.println(x+" "+y);
}else{
w.println("OPTIMAL");
}
}
w.close();
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 67e548489bcb182bbf05be36a10d3698 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
int saif = 13;
//hehe
Scanner ab=new Scanner(System.in);
int tc=ab.nextInt();
while(tc-->0)
{
int ss=ab.nextInt();
int min=(int)1e6;int ind=-1;
LinkedHashSet<Integer>set=new LinkedHashSet<>();
LinkedList<Integer>li=new LinkedList<>();
LinkedHashSet<Integer>set2=new LinkedHashSet<>();
for(int y=0;y<ss;y++)
{
int z=ab.nextInt();
int arr[]=new int[z];boolean fl=true;;
for(int i=0;i<z;i++)
{
arr[i]=ab.nextInt();
}
Arrays.sort(arr);
for(int a=0;a<z;a++)
{
if(set.contains(arr[a]))continue;
else if(fl&&!(set.contains(arr[a]))){set.add(arr[a]);li.addLast(arr[a]);set2.add(y);break;}
}
if(!(set2.contains(y)))ind=y;
}
if(set.size()==ss)System.out.println("OPTIMAL");
else
{
int i=0;
for(int j=1;j<=ss;j++){if(!(set.contains(j))){i=j;break;}}
System.out.println("IMPROVE");
System.out.println(ind+1+" "+i);
}
}
//#harouious#\\
}
} | Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | fbd67c1344dff90594f294ebf54741ff | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class q2
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc = new Scanner(System.in);
int t= Integer.parseInt(sc.next());
while(t-->0)
{
int n =Integer.parseInt(sc.next());
boolean[] pr=new boolean[n];
boolean[] ki= new boolean[n];
List<int[]> a1 = new ArrayList<>();
for(int i=0;i<n;i++)
{
int size= Integer.parseInt(sc.next());
int[] arr = new int[size];
for(int j=0;j<size;j++)
{
arr[j]= Integer.parseInt(sc.next());
}
a1.add(arr);
}
int count=0;
for(int[] p:a1)
{
for(int h:p)
{
if(!ki[h-1])
{
ki[h-1]=true;
pr[count]=true;
break;
}
}
count++;
}
List<Integer> vp= new ArrayList<>();
List<Integer> vq= new ArrayList<>();
int flag=0;
for(int i=0;i<n;i++)
{
if(!pr[i])
{
flag=1;
vp.add(i);
}
if(!ki[i])
{
vq.add(i);
}
}
if(flag==0)
{
System.out.println("OPTIMAL");
}
else{
int az= vq.size()<vp.size()?vq.size():vp.size();
System.out.println("IMPROVE");
for(int i=0;i<az;i++)
{
System.out.println((vp.get(i)+1)+" "+(vq.get(i)+1));
break;
}
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 8e7fed76d9df32dbe5a5389ee29b4e85 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
public class BigPP {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int l = -1;
ArrayList<Boolean> m = new ArrayList<Boolean>();
for(int i=0;i<n;i++)
m.add(false);
boolean fin = false;
for(int i=0;i<n;i++) {
int k = sc.nextInt();
boolean key = false;
for(int j=0;j<k;j++) {
int o = sc.nextInt();
if(!m.get(o - 1) && !key) {
m.set(o - 1, true);
key = true;
}
}
if(!key && !fin) {
l = i;
fin = true;
}
}
if(l == -1)
System.out.println("OPTIMAL");
else {
System.out.println("IMPROVE");
int index = 0;
for(int x=0;x<n;x++) {
if(!m.get(x)) {
index = x;
break;
}
}
System.out.println((l+1) +" " + (index+1));
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 3ac9c2d82bdb394ff77f6a286c2bc746 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
import java.util.Set;
public class code12 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0) {
int n = scn.nextInt();
HashMap<Integer, Integer> map = new HashMap<>();
for(int i=1; i<=n;i++) {
map.put(i, 1);
}
int a = -1;
for(int i=0; i<n; i++) {
int k = scn.nextInt();
if(k == 0)
a = i;
int flag = 0;
for(int j=0; j<k; j++) {
int b = scn.nextInt();
if(flag == 0 && map.containsKey(b)) {
map.remove(b);
flag = 1;
}
}
if(flag == 0)
a = i;
}
if(a == -1 || map.isEmpty())
System.out.println("OPTIMAL");
else {
System.out.println("IMPROVE");
System.out.print(a+1 + " ");
Set<Integer> key = map.keySet();
for(Integer k : key) {
System.out.println(k);
break;
}
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 0d525e576d99444734ded2531fe0dd93 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.util.*;
import java.io.*;
public class graph1 {
public static void main(String args[])throws IOException {
Scanner s=new Scanner(System.in);
int mod=998244353;
int t=s.nextInt();
while(t-->0) {
int n=s.nextInt();
int[] vis=new int[n+1];
ArrayList<Integer> jj[]=new ArrayList[n+1];
for (int i = 0; i < jj.length; i++) {
jj[i]=new ArrayList<>();
}
for(int i=1;i<=n;i++) {
int k=s.nextInt();
for(int j=0;j<k;j++) {
int rt=s.nextInt();
jj[i].add(rt);
}
}
int marr=0;
int girl=0;
int boy=0;
boolean flag=false;
for(int i=1;i<=n;i++) {
int j;
for( j=0;j<jj[i].size();j++) {
if(vis[jj[i].get(j)]==0) {
vis[jj[i].get(j)]=1;
marr++;
break;
}
}
if(j==jj[i].size() && flag==false) {
flag=true;
girl=i;
}
}
if(flag && marr!=n) {
for (int i = 1; i <n+1; i++) {
if(vis[i]==0) {
boy=i;break;
}
}
System.out.println("IMPROVE");
System.out.println(girl+" "+boy);
}else if(marr==n) {
System.out.println("OPTIMAL");
}
}
} }
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 017c79ef0faf7299e1af6bff78803261 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
public class Princesses_and_Princes {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-->0) {
int n = s.nextInt();
ArrayList<ArrayList<Integer>> al= new ArrayList<>();
for(int i =0;i<n;i++) {
int x = s.nextInt();
ArrayList<Integer> a1 = new ArrayList<>();
for(int j =0;j<x;j++) {
a1.add(s.nextInt());
}
al.add(a1);
}
boolean[] princes = new boolean[n];
boolean[] free = new boolean[n];
for(int i =0;i<n;i++) {
ArrayList<Integer> x = al.get(i);
for(int p:x) {
if(!free[p-1]) {
free[p-1] = true;
princes[i] = true;
break;
}
}
}
boolean flag = false;
for(int i =0;i<n;i++) {
if(!princes[i]) {
ArrayList<Integer> x = al.get(i);
for(int j =0;j<n;j++) {
if(!free[j] && !x.contains(j+1)) {
flag = true;
System.out.println("IMPROVE");
System.out.println((i+1)+" "+(j+1));
break;
}
}
if(flag) {
break;
}
}
}
if(!flag) {
System.out.println("OPTIMAL");
}
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 4216c1ca4c92cda589b862f49a8ff363 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes |
import java.io.*;
import java.util.*;
public class Princesses {
public static void main(String[] args) {
FastReader in = new FastReader();
int numTrials = in.nextInt();
while(numTrials --> 0) {
int numDaughters = in.nextInt();
ArrayDeque<Integer>[] daughterList = new ArrayDeque[numDaughters];
for(int i = 0; i < daughterList.length; i++) {
daughterList[i] = new ArrayDeque<Integer>();
int numKings = in.nextInt();
while(numKings --> 0) {
daughterList[i].addLast(in.nextInt());
}
}
int leftDaughter = -1;
HashSet<Integer> left = new HashSet<Integer>();
for(int i = 1; i <= numDaughters; i++) {
left.add(i);
}
for(int i = 0; i < daughterList.length; i++) {
boolean found = false;
ArrayDeque<Integer> curr = daughterList[i];
while(!curr.isEmpty()) {
int currHubby = curr.poll();
if(left.contains(currHubby)) {
left.remove(currHubby);
found = true;
break;
}
}
if(!found) {
leftDaughter = i + 1;
}
}
int leftPrince = -1;
if(left.size() >= 1) {
Iterator<Integer> iter = left.iterator();
leftPrince = iter.next();
}
if(leftDaughter != -1) {
System.out.println("IMPROVE");
System.out.println(leftDaughter + " " + leftPrince);
}
else {
System.out.println("OPTIMAL");
}
}
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 9a8e9f86f65be934be3d8d788b9e9956 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class B {
public static void main (String[] args) throws java.lang.Exception {
new Solution();
}
}
class Solution {
Reader reader;
public Solution() {
reader = new Reader();
int t = reader.ni();
while (reader.hasNext()) {
run_case();
}
}
private void run_case() {
int n = reader.ni();
// TreeMap<Integer, Integer> used = new TreeMap<>();
// int max_idx_prince = 0;
// for (int i=1; i<=n; i++) used.put(i, -1);
int[] arr_m = new int[n];
int[] arr_p = new int[n];
// int fail_married_idx = -1;
int cnt = 0;
for (int i=0; i<n; i++) {
int m = reader.ni();
boolean done = false;
for (int j=0; j<m; j++) {
int t = reader.ni();
if (!done && arr_p[t - 1] == 0) {
arr_m[i] = 1;
arr_p[t-1] = 1;
cnt++;
done = true;
}
}
}
if (cnt == n) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
int a = 0, b = 0;
for (int i=0; i<n; i++) {
if (arr_m[i] == 0) {
a = i+1;
break;
}
}
for (int i=0; i<n; i++) {
if (arr_p[i] == 0) {
b = i+1;
break;
}
}
System.out.println(a + " " + b);
}
return;
}
}
class Reader {
BufferedReader br;
StringTokenizer st;
public Reader(){
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
}
public String next() {
if (st.hasMoreTokens()) return st.nextToken();
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
return st.nextToken();
}
public boolean hasNext() {
try {
if (!st.hasMoreTokens() && !br.ready()) return false;
else return true;
} catch (Exception e) {
e.printStackTrace();
}
return false;
}
public int ni() {return Integer.parseInt(this.next());}
public String ns() {return this.next();}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | b25416adb9151f0df202356fb2f07b1e | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class B {
public static void main (String[] args) throws java.lang.Exception {
new Solution();
}
}
class Solution {
Reader reader;
public Solution() {
reader = new Reader();
int t = reader.ni();
while (reader.hasNext()) {
run_case();
}
}
private void run_case() {
int n = reader.ni();
TreeMap<Integer, Integer> used = new TreeMap<>();
// int max_idx_prince = 0;
for (int i=1; i<=n; i++) used.put(i, -1);
int fail_married_idx = -1;
for (int i=0; i<n; i++) {
int m = reader.ni();
boolean done = false;
for (int j=0; j<m; j++) {
int t = reader.ni();
if (used.containsKey(t) && !done) {
used.remove(t);
done = true;
}
}
if (!done) fail_married_idx = i + 1;
}
if (fail_married_idx == -1) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
System.out.println(fail_married_idx + " " + used.firstKey());
}
return;
}
}
class Reader {
BufferedReader br;
StringTokenizer st;
public Reader(){
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
}
public String next() {
if (st.hasMoreTokens()) return st.nextToken();
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
return st.nextToken();
}
public boolean hasNext() {
try {
if (!st.hasMoreTokens() && !br.ready()) return false;
else return true;
} catch (Exception e) {
e.printStackTrace();
}
return false;
}
public int ni() {return Integer.parseInt(this.next());}
public String ns() {return this.next();}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | c411dbbfedb8f008acb40b965f42e3ae | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class B {
public static void main (String[] args) throws java.lang.Exception {
new Solution();
}
}
class Solution {
Reader reader;
public Solution() {
reader = new Reader();
int t = reader.ni();
while (reader.hasNext()) {
run_case();
}
}
private void run_case() {
int n = reader.ni();
// TreeMap<Integer, Integer> used = new TreeMap<>();
// int max_idx_prince = 0;
// for (int i=1; i<=n; i++) used.put(i, -1);
int[] arr_m = new int[n];
int[] arr_p = new int[n];
// int fail_married_idx = -1;
int cnt = 0;
for (int i=0; i<n; i++) {
int m = reader.ni();
boolean done = false;
for (int j=0; j<m; j++) {
int t = reader.ni();
if (!done && arr_p[t - 1] == 0) {
arr_m[i] = 1;
arr_p[t-1] = 1;
cnt++;
done = true;
}
}
}
if (cnt == n) {
System.out.println("OPTIMAL");
} else {
System.out.println("IMPROVE");
int a = 0, b = 0;
for (int i=0; i<n; i++) if (arr_m[i] == 0) a = i+1;
for (int i=0; i<n; i++) if (arr_p[i] == 0) b = i+1;
System.out.println(a + " " + b);
}
return;
}
}
class Reader {
BufferedReader br;
StringTokenizer st;
public Reader(){
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
}
public String next() {
if (st.hasMoreTokens()) return st.nextToken();
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
return st.nextToken();
}
public boolean hasNext() {
try {
if (!st.hasMoreTokens() && !br.ready()) return false;
else return true;
} catch (Exception e) {
e.printStackTrace();
}
return false;
}
public int ni() {return Integer.parseInt(this.next());}
public String ns() {return this.next();}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 4f438ac0d4b73549bd37eb72abb6ea1c | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int T = sc.nextInt();
while(T-->0){
int n = sc.nextInt();
HashSet<Integer> set = new HashSet<>();
int ar[] = new int[n];
for(int i = 0; i<n; i++){
int count = sc.nextInt();
int min = Integer.MAX_VALUE;
while(count-->0){
int val = sc.nextInt();
if(set.contains(val))
continue;
min = Math.min(min,val);
}
if(min == Integer.MAX_VALUE){
ar[i] = -1;
}else{
set.add(min);
ar[i] = min;
}
}
int available=0;
for(int i =1; i<=n; i++){
if(!set.contains(i)){
available = i;
break;
}
}
if(available==0){
System.out.println("OPTIMAL");
}else{
System.out.println("IMPROVE");
for(int i=0; i<ar.length; i++){
if(ar[i]==-1){
System.out.println(i+1+" "+available);
break;
}
}
}
}
}
static final Random random = new Random();
static void sort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
int cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static void sortl(long[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
long cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 1293d73897b5c477ddb4b0a267a3302f | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class Princess_And_Princes {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int T = sc.nextInt();
while(T-->0){
int n = sc.nextInt();
// Optimal solution
HashSet<Integer> set = new HashSet<>();
int q = -1;
for (int i = 1; i <= n; i++) {
int k = sc.nextInt();
boolean flag = false;
while(k-->0) {
int temp = sc.nextInt();
if (!set.contains(temp) && !flag) {
flag = true;
set.add(temp);
}
}
if (!flag && q == -1) {
q = i;
}
}
if(q == -1){
System.out.println("OPTIMAL");
}else{
System.out.println("IMPROVE");
int king = 1;
while(king<=n){
if(!set.contains(king))
break;
king++;
}
System.out.println(q + " " + king);
}
}
}
static final Random random = new Random();
static void sort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
int cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static void sortl(long[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
long cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 0bcf8162f68a5752389b321daaff0558 | train_002.jsonl | 1584974100 | The King of Berland Polycarp LXXXIV has $$$n$$$ daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are $$$n$$$ other kingdoms as well.So Polycarp LXXXIV has enumerated his daughters from $$$1$$$ to $$$n$$$ and the kingdoms from $$$1$$$ to $$$n$$$. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the $$$n$$$-th daughter.For example, let there be $$$4$$$ daughters and kingdoms, the lists daughters have are $$$[2, 3]$$$, $$$[1, 2]$$$, $$$[3, 4]$$$, $$$[3]$$$, respectively. In that case daughter $$$1$$$ marries the prince of kingdom $$$2$$$, daughter $$$2$$$ marries the prince of kingdom $$$1$$$, daughter $$$3$$$ marries the prince of kingdom $$$3$$$, leaving daughter $$$4$$$ nobody to marry to.Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.Polycarp LXXXIV wants to increase the number of married couples.Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.For your and our convenience you are asked to answer $$$t$$$ independent test cases. | 256 megabytes | import java.io.*;
import java.util.*;
public class B {
public static void main(String[] args) {
FastScanner sc = new FastScanner();
int T = sc.nextInt();
while(T-->0){
int n = sc.nextInt();
HashSet<Integer> set = new HashSet<>();
int ar[] = new int[n];
for(int i = 0; i<n; i++){
int count = sc.nextInt();
int min = Integer.MAX_VALUE;
while(count-->0){
int val = sc.nextInt();
if(set.contains(val))
continue;
min = Math.min(min,val);
}
if(min == Integer.MAX_VALUE){
ar[i] = -1;
}else{
set.add(min);
ar[i] = min;
}
}
Queue<Integer> available = new LinkedList<>();
for(int i =1; i<=n; i++){
if(!set.contains(i))
available.add(i);
}
int ans[] = new int[available.size()];
for(int i = 0; i<ans.length; i++){
ans[i] = available.poll();
}
if(ans.length==0){
System.out.println("OPTIMAL");
}else{
System.out.println("IMPROVE");
for(int i=0; i<ar.length; i++){
if(ar[i]==-1){
System.out.println(i+1+" "+ans[0]);
break;
}
}
}
}
}
static final Random random = new Random();
static void sort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
int cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static void sortl(long[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int val = random.nextInt(n);
long cur = a[i];
a[i] = a[val];
a[val] = cur;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
| Java | ["5\n4\n2 2 3\n2 1 2\n2 3 4\n1 3\n2\n0\n0\n3\n3 1 2 3\n3 1 2 3\n3 1 2 3\n1\n1 1\n4\n1 1\n1 2\n1 3\n1 4"] | 2 seconds | ["IMPROVE\n4 4\nIMPROVE\n1 1\nOPTIMAL\nOPTIMAL\nOPTIMAL"] | NoteThe first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.In the second test case any new entry will increase the number of marriages from $$$0$$$ to $$$1$$$.In the third and the fourth test cases there is no way to add an entry.In the fifth test case there is no way to change the marriages by adding any entry. | Java 8 | standard input | [
"greedy",
"graphs",
"brute force"
] | 38911652b3c075354aa8adb2a4c6e362 | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^5$$$) — the number of test cases. Then $$$t$$$ test cases follow. The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of daughters and the number of kingdoms. Each of the next $$$n$$$ lines contains the description of each daughter's list. The first integer $$$k$$$ ($$$0 \le k \le n$$$) is the number of entries in the $$$i$$$-th daughter's list. After that $$$k$$$ distinct integers follow $$$g_i[1], g_i[2], \dots, g_i[k]$$$ ($$$1 \le g_i[j] \le n$$$) — the indices of the kingdoms in the list in the increasing order ($$$g_i[1] < g_i[2] < \dots < g_i[k]$$$). It's guaranteed that the total number of daughters over all test cases does not exceed $$$10^5$$$. It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed $$$10^5$$$. | 1,200 | For each test case print the answer to it. Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list. If there are multiple ways to add an entry so that the total number of married couples increases then print any of them. Otherwise the only line should contain one word "OPTIMAL". | standard output | |
PASSED | 1a6730e62f00ccd947dc9addcf9b3808 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class CF344A_Magnets {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int count = 1;
boolean curPos = false;
for (int i = 0; i < n; i++) {
boolean pos = readLine(in).startsWith("1");
if (i == 0)
curPos = pos;
else {
if (curPos != pos) {
count++;
curPos = pos;
}
}
}
System.out.println(count);
}
private static String readLine(Scanner in) {
String line = "";
while ("".equals(line))
line = in.nextLine();
return line;
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 71bcdd696831f18e01acc0f31f18f3c4 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Magnets {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int nr = sc.nextInt();
int x, old = 0, groups = 0;
for(int i = 0; i < nr; i++) {
x = sc.nextInt();
if(x != old) {
groups++;
}
old = x;
}
System.out.println(groups);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 8f448f5633d5a5ddc12e86c90156838b | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class A {
BufferedReader in;
StringTokenizer st;
PrintWriter out;
public void Solution() throws IOException {
int n = nextInt();
int ans = 0;
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
for (int i = 1; i < n; i++) {
if (a[i] != a[i - 1]) {
ans++;
}
}
System.out.println(ans + 1);
}
public void run() throws IOException {
try {
// in = new BufferedReader(new FileReader(new File("input.txt")));
// out = new PrintWriter(new File("output.txt"));
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
Solution();
out.close();
} catch (Exception e) {
e.printStackTrace();
System.exit(261);
}
}
public String nextToken() throws IOException {
if (st == null || !st.hasMoreTokens()) {
st = new StringTokenizer(in.readLine());
}
return st.nextToken();
}
public long nextlong() throws IOException {
return Long.parseLong(nextToken());
}
public int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static void main(String args[]) throws IOException {
new A().run();
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 79cf794abfc70aafdffc322d3932b48b | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Main{
public static void main(String [] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[]= new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t=1;
int temp=a[0];
for(int i=1;i<n;i++){
if(temp!=a[i]){
temp=a[i];
t++;
}
}
System.out.print(t);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 0152cf1ceac278b4d42e04ae6f4c6348 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class NumberOne {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
int islands = 1;
for (int i = 1; i < n; i++) {
if (a[i - 1] != a[i]) {
islands++;
}
}
System.out.println(islands);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | c987393733d5c75953e6edcff2719c5f | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
public class Main
{
public static void main(String[] args) throws IOException
{
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
//BufferedReader input = new BufferedReader(new FileReader("Sample Input.txt"));
//StringBuilder output = new StringBuilder();
int n = Integer.parseInt(input.readLine());
int groups=0;
String line = input.readLine();;
String newLine;
for (int i = 0; i < n; i++)
{
if (line.equals(newLine=input.readLine()))
{
line = newLine;
}
else
{
groups++;
line = newLine;
}
}
System.out.println(groups);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 1808879b7c7b77faf07efd23409e894c | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A344 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
int acc=0;
int prev=0;
for(int i=0;i<N;i++) {
int x = in.nextInt();
if(prev!=x) {
prev=x;
acc++;
}
}
System.out.println(acc);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | c4565fd35396d9403626812aeebde0fa | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
public class task344a {
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
try {
int n = Integer.parseInt(br.readLine());
int count = 0;
String cur = "ss";
String str;
for (int i = 0; i < n; i++) {
str = br.readLine();
if (str.compareTo(cur) != 0) {
count++;
cur = str;
}
}
System.out.println(count);
} catch (IOException e) {
}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 9f412c4c1059a39b47e99d70e742d20f | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class ContestTeam1 {
/**
* @param args
*/
public static void main(String[] args) {
Scanner console = new Scanner(System.in);
int count = 0;
while (console.hasNext()) {
int n = console.nextInt();
String prevNum = console.next();
count = 1;
for (int i=1;i<n;i++) {
String num = console.next();
if(prevNum.charAt(1)==num.charAt(0))
{
count++;
}
prevNum = num;
}
System.out.println(count);
}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 94a32896ce2b8d5c5a8225752916d61b | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A344 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
String []dominoes = new String[n];
for(int i = 0; i < n; ++i){
dominoes[i] = scanner.next();
}
int count = 1;
for(int i = 1; i < n; ++i){
if(dominoes[i].equals(dominoes[i-1])){
continue;
}else {
count ++;
}
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 5a7de80bcc6fa25caaca70885ddaab84 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
String str = "";
int count = 0;
for(int i=0; i<n; i++) {
String tmp = s.next();
if(!tmp.equals(str)) {
str = tmp;
count++;
}
}
System.out.println(count);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 55f5619ba16bdf2323fe2eaa2d5b9f7c | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int numOfMang = sc.nextInt();
int []magPos = new int[100000];
for(int i =0;i<numOfMang;++i)
{
magPos[i] = sc.nextInt();
}
int count = 0;
for(int i =1;i<numOfMang;++i){
if(magPos[i]==magPos[i-1])
continue;
else
count+=1;
}
System.out.println(count+1);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 5bd54ec7a76e4fac9f26ff010cca1f89 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int count = 1;
int x = in.nextInt(); int temp;
while(n>1) {
temp = x;
x = in.nextInt();
if(temp!=x) count++;
n--;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 7614df98e612716b28a09b73d04de94b | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Magnets {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
// while (scanner.hasNext()) {
int n = scanner.nextInt();
int old = 0;
int count = 0;
int maxCount = 0;
for (int i = 0; i < n; i++) {
int a = scanner.nextInt();
if(a == old)
count ++;
else
{
old = a;
maxCount ++;
}
}
System.out.println(maxCount);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 0cc3c43fc2f2e2a9e3f03b45db9b888d | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes |
import java.util.Scanner;
public class A344 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int count = 1;
scan.nextLine();
String str1 = scan.nextLine();
String str2 = "";
for (int i = 1; i < n; i++) {
str2 = scan.nextLine();
if (str2.charAt(0) == str1.charAt(1)) {
count++;
}
str1 = str2;
}
System.out.println(count);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 4b26853987c7588bd9aaa33224b3509a | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n = Integer.parseInt(cin.nextLine());
String c = "";
int r = 0;
for (int i = 0; i < n; i++) {
String input = cin.nextLine();
if (!input.equals(c))
r++;
c = input;
}
System.out.println(r);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 460e6965905370d28424c26a8bbf0c3b | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String args[]) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = t_int(br);
StringBuilder out = new StringBuilder();
while (n-- > 0) {
out.append(br.readLine());
}
char[] magnets = out.toString().toCharArray();
int v = 1;
for (int i = 0; i < magnets.length - 1; i++) {
if (magnets[i] == magnets[i + 1]) {
v++;
}
}
System.out.println(v);
br.close();
}
public static int t_int(BufferedReader br) throws Exception {
return Integer.parseInt(br.readLine());
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 9d4b12948d77c604302f15a04a6ec33a | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
while(cin.hasNext()){
int numberOfMagnets = cin.nextInt();
char ultimoSigno='a';
int contador = 1;
for(int i=0;i<numberOfMagnets;i++){
String magneto = cin.next();
if(ultimoSigno == 'a'){
ultimoSigno = magneto.charAt(1);
}else {
if(ultimoSigno == magneto.charAt(0)){
contador++;
ultimoSigno = magneto.charAt(1);
}else{
ultimoSigno = magneto.charAt(1);
}
}
}
System.out.println(contador);
}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 3b8e097e1e5cc8555afe1ed89572ff0f | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Arrays;
import java.util.Scanner;
public class Main{
private static Scanner in = new Scanner(System.in);
private static int task;
private static String[] stArr;
private static int[] inArr;
private static void solve() {
int z = Integer.parseInt(in.nextLine());
int cou = 1;
String[] dom = in.nextLine().split("");
String[] pol;
for (int i = 0; i < z - 1; i++) {
pol = in.nextLine().split("");
if (pol[1].equals(dom[2])) {
cou++;
}
dom = pol;
}
System.out.println(cou);
}
public static void main(String[] args) {
// task = in.nextInt();
new Thread(new Dev()).start();
}
public static class Dev implements Runnable {
@Override
public void run() {
// while (task > 0) {
solve();
// --task;
// }
}
}
public static void view(Object... os) {
System.err.println(Arrays.deepToString(os));
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | a24a168902c189a9c696dc03f19e5f58 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int z = Integer.parseInt(in.nextLine());
int cou = 1;
String[] dom = in.nextLine().split("");
String[] pol;
for (int i = 0; i < z - 1; i++) {
pol = in.nextLine().split("");
if (pol[1].equals(dom[2])) {
cou++;
}
dom = pol;
}
System.out.println(cou);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 0bfec6d802016ced864ddc25aad08c7d | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.PrintWriter;
import java.util.Scanner;
public class n344A {
Scanner in;
PrintWriter out;
void solve() {
int n = in.nextInt();
in.nextLine();
StringBuilder sb = new StringBuilder("");
for (int i = 0; i < n; i++) {
sb.append(in.nextLine());
}
int ans = 1;
for (int i = 2; i < sb.length(); i += 2) {
if (sb.charAt(i) == sb.charAt(i - 1)) {
ans++;
}
}
out.println(ans);
}
void run() {
in = new Scanner(System.in);
out = new PrintWriter(System.out);
try {
solve();
} finally {
out.close();
}
}
public static void main(String[] args) {
new n344A().run();
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | b4ba6117eca422ed893e3ff7239056ce | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int x = input.nextInt();
String s; int num = 1;
char[] c = new char[x+1];
for(int i = 1; i<= x; i++)
{
s = input.next();
c[i] =s.charAt(1);
}
for(int i = 1; i< c.length; i++)
{
if(i != c.length-1)
{
if(c[i] != c[i+1])
num++;
}
}
System.out.println(num);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 698ef61075cc5e1b9c778763976de60e | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
/*
br = new BufferedReader(new FileReader("input.txt"));
pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
*/
public class Main {
private static BufferedReader br;
private static StringTokenizer st;
private static PrintWriter pw;
public static void main(String[] args) throws IOException {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
//int qq = 1;
int qq = Integer.MAX_VALUE;
//int qq = readInt();
for(int casenum = 1; casenum <= qq; casenum++) {
int n = readInt();
String[] list = new String[n];
for (int i = 0; i < list.length; i++) {
list[i] = nextToken();
}
int ret = 1;
for(int i = 1; i < n; i++) {
if(!list[i].equals(list[i-1])) {
ret++;
}
}
pw.println(ret);
}
pw.close();
}
private static void exitImmediately() {
pw.close();
System.exit(0);
}
private static long readLong() throws IOException {
return Long.parseLong(nextToken());
}
private static double readDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private static int readInt() throws IOException {
return Integer.parseInt(nextToken());
}
private static String nextToken() throws IOException {
while(st == null || !st.hasMoreTokens()) {
if(!br.ready()) {
exitImmediately();
}
st = new StringTokenizer(br.readLine().trim());
}
return st.nextToken();
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 13b902ad88017563aa68aedffa124419 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
public class taskA {
Scanner in;
PrintWriter out;
void solve() {
int n = in.nextInt();
int one = in.nextInt();
int two;
int col = 1;
for (int i = 1; i < n; i++) {
two = in.nextInt();
if(one != two){
col++;
}
one = two;
}
out.println(col);
}
void run() {
in = new Scanner(System.in);
out = new PrintWriter(System.out);
try {
solve();
} finally {
out.close();
}
}
public static void main(String[] args) {
new taskA().run();
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | fb79ec45437914bfaf13b0045287a125 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/**
*
* @author jupiter
*/
import java.util.Scanner;
import java.math.BigInteger;
import java.util.Arrays;
public class main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
sc.nextLine();
String[] s = new String[N];
for(int i=0; i<N; i++) {
s[i] = sc.nextLine();
}
int count = 1;
for(int i=0; i<N-1; i++) {
if(!s[i].equals(s[i+1])) count++;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 7d8c7a9640971f9d5eaa3dfc8c799d1e | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int N =Integer.parseInt(in.readLine());
int cant = 1;
int ant = -1;
for(int i=0;i<N;++i){
String ln = in.readLine();
if(ant!=-1){
int p = ln.charAt(0)-'0';
if(ant==p)
cant++;
}
ant = ln.charAt(1)-'0';
}
System.out.println(cant);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | df59d9db61fd7b8fdc0404748d7bffb0 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
public class Main{
public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
t--;
int d=1;
int a=Integer.parseInt(br.readLine());
while(t-->0){
int b=Integer.parseInt(br.readLine());
if(a!=b){
a=b;
d++;
}
}
System.out.println(d);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 038c0210ff5c84fcf6ce8d0b8c702dc3 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class CF200_Div2_A {
static Scanner sc=new Scanner(System.in);
public static void main(String[] args){
int n=sc.nextInt();
int ans=1;
String[] a=new String[n];
for(int i=0;i<n;i++)a[i]=sc.next();
for(int i=0;i<n-1;i++)if(!a[i].equals(a[i+1]))ans++;
System.out.println(ans);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 698aa0481d950626b26ffb28299f2461 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A {
/**
* @param args
*/
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int prev = in.nextInt();
int count = 1;
for (int i = 1; i < n; i++) {
int m = in.nextInt();
if (m != prev) {
count++;
}
prev = m;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 6f480d42f65e8ac3b8fd08382feaf90a | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class main
{
public static void main(String[] args) throws IOException
{
Scanner in = new Scanner(System.in);
int ans = 0;
int n = in.nextInt();
String prev = "";
for(int i = 0; i < n; i++)
{
String cur = in.next();
if(cur.compareTo(prev) != 0)
{
prev = cur;
ans++;
}
}
System.out.println(ans);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 23f7a75363871ba05ed0fd75fcf71568 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
Scanner e=new Scanner(System.in);
int n=e.nextInt();
int sw=0;
String aux="00";
int c=0;
e.nextLine();
for (int i = 0; i < n; i++)
{
String cad=e.nextLine();
//System.out.println("** "+i+" "+cad+" * "+aux+" "+c);
if(sw==0)
{
sw=1;
aux=cad;
c++;
}
else
{
if(!cad.equals(aux))
{
//System.out.println("* "+cad+" "+aux);
aux=cad;
c++;
}
}
}
System.out.println(c);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 7ed7c591d93b525876c51296b6c4f984 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | /**
* @(#)Magnets.java
*
*
* @author
* @version 1.00 2013/10/17
*/
import java.util.Scanner;
public class Magnets {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int k = 0;
int j;
int count = 0;
for (int i = 0; i < n; i++){
j = input.nextInt();
if (j != k)
count++;
k = j;
}
System.out.println(count);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | ecab751faee447d7cd51faaf16066cda | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Magnets {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt(); s.nextLine();
StringBuilder str = new StringBuilder("");
for(int z = 0; z < n; z++){
str.append(s.nextLine());
}
int res = 1;
for(int a = 1; a < str.length(); a++){
if(str.charAt(a) == str.charAt(a - 1)) res++;
}
System.out.println(res);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 33bdcfe5c57886b4c1e633bf5b1e8f96 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A344 {
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
String prev = "";
int ret = 0;
int n = in.nextInt();
for (int i = 0; i < n; i++)
{
String cur = in.next();
if (!prev.equals(cur))
ret++;
prev = cur;
}
System.out.println(ret);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 212fc7549e75997a0dfeefa74a7506cc | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.PrintWriter;
import java.util.Scanner;
public class A {
static Scanner in; static int next() throws Exception {return in.nextInt();};
// static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;}
// static BufferedReader in;
static PrintWriter out;
public static void main(String[] args) throws Exception {
in = new Scanner(System.in);
// in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
// in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
int n = next();
int res = 1;
char c[][] = new char[n][2];
for (int i = 0;i < n;i++){
c[i] = in.next().toCharArray();
}
for (int i = 1;i < n;i++) {
if (c[i-1][0] != c[i][0]) res++;
}
out.println(res);
out.println();
out.close();
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 0444663d076bfe35e686017ea253de95 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class A implements Runnable {
private void solve() throws IOException {
int n = nextInt();
String[] s = new String[n];
for (int i = 0; i < n; i++) s[i] = nextToken();
int ans = 1;
for (int i = 1; i < n; i++) if (s[i].charAt(0) == s[i-1].charAt(1)) ans++;
print(ans);
}
public static void main(String[] args) {
new A().run();
}
BufferedReader reader;
StringTokenizer tokenizer;
PrintWriter writer;
public void run() {
try {
reader = new BufferedReader(new InputStreamReader(System.in));
writer = new PrintWriter(System.out);
solve();
writer.close();
} catch (Exception e) {
e.printStackTrace();
System.exit(261);
}
}
void halt() {
writer.close();
System.exit(0);
}
void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
void println(Object... objects) {
print(objects);
writer.println();
}
String nextLine() throws IOException {
return reader.readLine();
}
String nextToken() throws IOException {
while (tokenizer == null || !tokenizer.hasMoreTokens())
tokenizer = new StringTokenizer(nextLine());
return tokenizer.nextToken();
}
int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(nextToken());
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | d93bedfade0bbe18171c09b8bd66e566 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Magnet {
static String pm = "01", mp = "10", bf = "00";
public static void main(String[] args) {
System.out.println(cal(args));
}
static int cal(String[] args) {
int group = 0;
Scanner in = new Scanner(System.in);
int num= in.nextInt();
if (num < 1 || num > 100000) {return 0;}
for (int i = 0; i < num; i++) {
String now = in.next();
if (!bf.equals(now)) {
bf = now;
group++;
}
}
return group;
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 38c1b307d68a299db9bc102eb3f67432 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class Magnets {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
try{
BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
PrintWriter p=new PrintWriter(System.out,true);
while(reader.ready()){
int n=Integer.parseInt(reader.readLine());
ArrayList <String> a=new ArrayList <String>();
while(n-->0){
a.add(reader.readLine());
}
int f=1;
for(int i=1;i<a.size();i++){
if(a.get(i).matches(a.get(i-1)))
continue;
else
f++;
}
p.println(f);
}
}
catch(Exception e){}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | aaeb88abf10fea0fc9e1ceabad6aaa0d | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
public class Magnets {
/**
* @param args
*/
public static void main(String[] args) throws IOException{
// TODO Auto-generated method stub
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
if(n>=1&&n<=100000){
String []a=new String[n];
int c=1;
for(int i=0;i<n;i++){
a[i]=br.readLine().trim();
if(i!=0){
if(!a[i].equals(a[i-1]))
c++;
}
}
System.out.println(c);
}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 6012f1f02dbc5d5b8f8767ae8c9ad14d | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import static java.lang.System.out;
import java.util.Scanner;
import java.lang.StringBuilder;
public class Main2{
public static void main(String[] args){
Scanner kb = new Scanner(System.in);
int n = kb.nextInt();
String[] s = new String[n];
s[0] = kb.next();
int cnt = 0;
for(int i = 1; i < n; i++){
s[i] = kb.next();
if(!s[i].equals(s[i-1]))
cnt++;
}
out.println(cnt+1);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 31c1140d092bd39754202448b2e0c388 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;
public class Magnits {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int numberDataInArray = sc.nextInt();
StringBuilder data = new StringBuilder();
int countZero = 0;
boolean isPrevZero = false;
for (int i = 0; i < numberDataInArray; i++) {
if(sc.next().equals("10")){
data.append("1");
isPrevZero = false;
}
else if(!isPrevZero){
countZero ++;
isPrevZero = true;
data.append("0");
}
}
System.out.println(trimArray(data.toString().split("0")).size() + countZero);
}
public static List<String> trimArray(String [] data){
List<String> result = new LinkedList<String>();
for (String currentData : data) {
if(!currentData.isEmpty()){
result.add(currentData);
}
}
return result;
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | a187dcfc8d9db778808cebc1ba93b1a5 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A {
public static void main(String[]args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
String[]array=new String[n];
for(int i=0;i<n;i++)
array[i]=sc.next();
int count=1;
for(int i=0;i<n-1;i++){
if(!array[i].equals(array[i+1]))
count++;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 4e078e1642a593609369ac4304951f3d | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class MagnetMike {
public static int t = 0;
public static BufferedReader in;
public static StringTokenizer st;
public static String nextToken() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(in.readLine());
return st.nextToken();
}
public static int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
public static void main(String[] args) throws Exception {
in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
int n= nextInt();
int g = 1;
String prev = nextToken();
for(int i=1;i<n;i++){
String mo = nextToken();
if(!mo.equals(prev))
{
g++;
prev = mo;
}
}
out.println(g);
out.flush();
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 9da0af0578c47ed91bc6a3e5707c89c2 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class A344 {
public static void main(String args[]) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
int count = 1;
for (int i = 1; i < n; i++) {
if (a[i] != a[i - 1])
++count;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 433cf23e49e294420538baf81383d1b8 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class Main{
public static void main(String[]args){
Scanner input=new Scanner(System.in);
int n=input.nextInt();int p=1;
int[]a=new int[n];
for(int i=0;i<n;i++){
a[i]=input.nextInt();
}
for(int i=0;i<n-1;i++){
if(a[i]==a[i+1]){continue;}
else{p++;}
}
System.out.println(p);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | e5270a9ff2a23f486371a92d78c76fda | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes |
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int [] x = new int[n];
x[0]=input.nextInt();
int count = 1 ;
for(int i=1 ;i<n; i++){
x[i]=input.nextInt();
if(x[i]!= x[i-1])
count ++ ;
}
System.out.println (count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | d81dddee6cf10ff6edbd637b8813e250 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
public class magnets {
public static void main(String[] agrs){
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int number = in.nextInt();
int endPre = number % 10;
int start;
int count = 1;
for (int i = 0; i < n -1 ; i++)
{
number = in.nextInt();
start = number/10;
if (endPre == start)
count++;
endPre = number % 10;
}
if (n == 0)
System.out.println(0);
else System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | e3a51ce472a61aea76f2e7db1d339928 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Problem344A {
public static void main(String[] args) throws Exception {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
String l=scanner.next();
int groups = 1;
for(int i=0; i<n-1; i++){
String m = scanner.next();
if(l.charAt(1)==m.charAt(0))
groups++;
l = m;
}
System.out.println(groups);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 21c616d08e405c6f88dd39eb6a3165bd | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int ans =0;
String temp ="";
while(n-->=0){
String s = in.nextLine();
if(!temp.equals(s)){
ans++;
}
temp = s;
}
System.out.println(ans);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 39969d98aa209849c0d43faf97c49fd0 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class JavaApplication1 {
public static void main(String[] args) {
Scanner scanner = new Scanner( System.in );
int n = scanner.nextInt();
int count=0;
String last=scanner.nextLine();
String temp;
for (int i=0;i<n;i++){
temp=scanner.nextLine();
if (!temp.equals(last)){
count++;
}
last=temp;
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | fd313b94bdf4d35f37768a08c36e96b0 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class Smile {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int ans = n;
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = in.nextInt();
}
for (int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
ans--;
}
}
out.println(ans);
out.close();
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 608169f9d5e2bec786fee73617e6c80c | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes |
import java.io.*;
import java.util.*;
public final class TestClass{
static ArrayList<String> arr1 = new ArrayList<String>();
static ArrayList<Integer> arr2 = new ArrayList<Integer>();
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
Scanner sc = new Scanner(System.in);
int n=Integer.parseInt(br.readLine());
while(n>0){
arr1.add(br.readLine());
n--;
}
if(arr1.size()==1){
System.out.println("1");
return;
}
int count=0;
for(int i=1;i<arr1.size();i++){
if(arr1.get(i).charAt(0)==arr1.get(i-1).charAt(1)){
count++;
}
}
System.out.println(count+1);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 5bf98eab870671e2872498deb3c1e407 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
public class Contest200A {
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
//BufferedReader in = new BufferedReader(new FileReader("200A.in"));
int count = 1;
char current = '9';
int n = Integer.parseInt(in.readLine());
for (int i = 0; i < n; i++) {
String s = in.readLine();
char a = s.charAt(0);
if (a == current) count++;
current = s.charAt(1);
}
System.out.println(count);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | bb8334d12513add45ae69f0324db9323 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.*;
import java.io.*;
public class cfa {
static long mod = 1000000009;
static class Pair1 implements Comparable<Pair1>
{
int a, b, i;
public Pair1(int aa, int bb, int ii)
{
a = aa; b = bb; i = ii;
}
@Override
public int compareTo(Pair1 o) {
// TODO Auto-generated method stub
if(this.b == o.b) return o.a - this.a;
return o.b - this.b;
}
}
static class Pair2 implements Comparable<Pair2>
{
int a, b, i;
public Pair2(int aa, int bb, int ii)
{
a = aa; b = bb; i = ii;
}
@Override
public int compareTo(Pair2 o) {
// TODO Auto-generated method stub
if(this.a == o.a) return o.b - this.b;
return o.a - this.a;
}
}
public static void main(String[] args) throws IOException {
// Scanner input = new Scanner(new File("input.txt"));
// PrintWriter out = new PrintWriter(new File("output.txt"));
input.init(System.in);
PrintWriter out = new PrintWriter((System.out));
int n = input.nextInt();
String s = "";
int res = 0;
for(int i = 0; i<n; i++)
{
String t = input.next();
if(!s.equals(t))res++;
s = t;
}
out.println(res);
out.close();
}
static long pow(long x, long p) {
if (p == 0)
return 1;
if ((p & 1) > 0) {
return (x * pow(x, p - 1)) % mod;
}
long sqrt = pow(x, p / 2);
return (sqrt * sqrt) % mod;
}
static long gcd(long a, long b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
static class input {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(new InputStreamReader(input));
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) {
// TODO add check for eof if necessary
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt(next());
}
static double nextDouble() throws IOException {
return Double.parseDouble(next());
}
static long nextLong() throws IOException {
return Long.parseLong(next());
}
static String nextLine() throws IOException {
return reader.readLine();
}
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | e7b844de48ac1223c1a8216d74870a7c | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.PrintWriter;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
String[] m = new String[n];
int cnt = 1;
for (int i = 0; i < n; i++) {
m[i] = in.next();
}
for (int i = 0; i < n - 1; i++) {
if (m[i].substring(1, 2).equals(m[i + 1].substring(0, 1))) {
cnt++;
}
}
out.println(cnt);
out.close();
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 6b0e6cd320ea52487120886caf6ea885 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class A {
/**
* @param args
* @throws IOException
*/
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
char c = in.readLine().charAt(1);
int groups = 1;
for (int i = 0; i < n-1; i++) {
String x = in.readLine();
if(x.charAt(0)== c){
groups++;
c = x.charAt(1);
}
}
System.out.println(groups);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 32e65c67e915160ce007d1c9216687ab | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int n , ans=1;
Scanner input = new Scanner(System.in);
n=input.nextInt();
String arr[]=new String[n];
for(int i=0 ; i<n ; i++){
String st;
st=input.next();
arr[i]=st;
}
if(arr.length>0){
for(int j=1 ; j<n ; j++){
if(arr[j].charAt(0)==arr[j-1].charAt(1) )
ans+=1;
}}
System.out.println(ans);
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 503cae218ff769394561eaa42a01d1f8 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.util.Scanner;
public class A {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans = 1;
int a = 0, b = 0;
b = sc.nextInt();
for (int i = 0; i < n-1; i++) {
a = sc.nextInt();
if(a != b){
ans++;
}
b = a;
}
System.out.println(ans);
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 9293e682552a8329e91341b695290e82 | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
import static java.lang.Math.*;
public class Solution implements Runnable {
public void solve() throws Exception {
int n = sc.nextInt();
int lst = -1;
int ans = 0;
while (n-- > 0){
int a = sc.nextInt();
if(a == 10 && lst == 1){
ans++;
}
if(a == 1 && lst == 0){
ans++;
}
lst = a%10;
}
out.println(ans + 1);
}
static Throwable throwable;
BufferedReader in;
PrintWriter out;
FastScanner sc;
public static void main(String[] args) throws Throwable {
Thread thread = new Thread(null, new Solution(), "", (1 << 26));
thread.start();
thread.join();
if (Solution.throwable != null)
throw throwable;
}
public void run() {
try {
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
sc = new FastScanner(in);
solve();
} catch (Throwable throwable) {
Solution.throwable = throwable;
} finally {
out.close();
}
}
}
class FastScanner {
BufferedReader reader;
StringTokenizer strTok;
FastScanner(BufferedReader reader) {
this.reader = reader;
}
public String nextToken() throws Exception {
while (strTok == null || !strTok.hasMoreTokens()) {
strTok = new StringTokenizer(reader.readLine());
}
return strTok.nextToken();
}
public int nextInt() throws Exception {
return Integer.parseInt(nextToken());
}
public long nextLong() throws Exception {
return Long.parseLong(nextToken());
}
public double nextDouble() throws Exception {
return Double.parseDouble(nextToken());
}
} | Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output | |
PASSED | 0f1028f05fc0d06411676689be5d0afd | train_002.jsonl | 1379172600 | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own. Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | 256 megabytes | import java.io.*;
import java.util.*;
public class _344_A implements Runnable {
private BufferedReader br = null;
private PrintWriter pw = null;
private StringTokenizer stk = new StringTokenizer("");
public static void main(String[] args) {
new Thread(new _344_A()).run();
// TODO Auto-generated method stub
}
public void run() {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new OutputStreamWriter(System.out));
solver();
pw.close();
}
public void nLine() {
try {
if (!stk.hasMoreTokens())
stk = new StringTokenizer(br.readLine());
} catch (Exception e) {
// TODO: handle exception
}
}
public String nstr() {
while (!stk.hasMoreTokens())
nLine();
return stk.nextToken();
}
public int ni() {
return Integer.valueOf(nstr());
}
public double nd() {
return Double.valueOf(nstr());
}
private void solver() {
int n = ni(), t =0;
String []s = new String[n];
int []ans = new int[n];
for (int i=0; i<n; i++)
{
s[i]=nstr();
if (i!=0)
{
if (s[i].charAt(0) != s[i-1].charAt(1))
{
ans[t] ++;
}
else
{
t++;
}
}
}
System.out.println((t+1));
}
}
| Java | ["6\n10\n10\n10\n01\n10\n10", "4\n01\n01\n10\n10"] | 1 second | ["3", "2"] | NoteThe first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.The second testcase has two groups, each consisting of two magnets. | Java 6 | standard input | [
"implementation"
] | 6c52df7ea24671102e4c0eee19dc6bba | The first line of the input contains an integer n (1 ≤ n ≤ 100000) — the number of magnets. Then n lines follow. The i-th line (1 ≤ i ≤ n) contains either characters "01", if Mike put the i-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | 800 | On the single line of the output print the number of groups of magnets. | standard output |
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