| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14363 | Accepted: 8048 |
Description
N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.
The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.
Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B
Output
* Line 1: A single integer representing the number of cows whose ranks can be determined
Sample Input
5 5 4 3 4 2 3 2 1 2 2 5
Sample Output
2
Source
USACO 2008 January Silver
import java.util.*;
public class Main {static int map[][];static int n,m;public static void floyd(){int i,j,k;for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++){if (map[i][j] == 0 && map[i][k]==1 && map[k][j]==1)//遍历所有传递关系map[i][j]=1;}}public static void main(String[] args) {// write your code hereScanner in=new Scanner(System.in);n=in.nextInt();m=in.nextInt();map=new int[n+1][n+1];for(int i=0;i<m;i++){int a=in.nextInt();int b=in.nextInt();map[a][b]=1;//如果a>b 则标记为1}floyd();int count=0;for(int i=1;i<=n;i++){int sum=0;for(int j=1;j<=n;j++){if(map[i][j]==1 || map[j][i]==1)//如果一头牛与其他n-1头牛都存在关系，则可确定排名{sum++;}}if(sum==n-1){count++;}}System.out.println(count);}}