POJ 3660 cow contest （Folyed 求传递闭包）

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
题解：
有N头牛，然后给你M个关系，
每个关系两头牛的编号：代表前者打败后者；
然后让你求可以确定名次的有多少头牛；
我们可以利用Folyed 来找找出每头牛可以和多少头牛建立关系，当且一头牛可以和剩下的N-1头牛都可以建立关系时，它的名次就可以确定了；求和即可达到答案；
参考代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<set>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
#define PI acos(-1)
#define EPS 1e-8
const int INF=0x3f3f3f3f;
const LL inf=0x3f3f3f3f3f3f3f3fLL;

const int maxn=110;
int N,M,A,B,dp[maxn][maxn];
int main()
{
    scanf("%d%d",&N,&M);
    memset(dp,0,sizeof dp);
    for(int i=1;i<=M;i++) scanf("%d%d",&A,&B),dp[A][B]=1;
    
    for(int k=1;k<=N;k++)
        for(int i=1;i<=N;i++)
            for(int j=1;j<=N;j++) if(dp[i][k]&&dp[k][j]) dp[i][j]=1;
            
    int ans=0,j;
    for(int i=1;i<=N;i++)
    {
        for(j=1;j<=N;j++) 
        {
            if(i==j) continue;
            if(!dp[i][j]&&!dp[j][i]) break;
        }
        if(j>N) ans++;
    }
    printf("%d
",ans);
    return 0;
}