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=;
 int N,M,A,B,dp[maxn][maxn];
 int main()
 {
     scanf("%d%d",&N,&M);
     memset(dp,,sizeof dp);
     for(int i=;i<=M;i++) scanf("%d%d",&A,&B),dp[A][B]=;

     for(int k=;k<=N;k++)
         for(int i=;i<=N;i++)
             for(int j=;j<=N;j++) if(dp[i][k]&&dp[k][j]) dp[i][j]=;

     int ans=,j;
     for(int i=;i<=N;i++)
     {
         for(j=;j<=N;j++)
         {
             if(i==j) continue;
             if(!dp[i][j]&&!dp[j][i]) break;
         }
         if(j>N) ans++;
     }
     printf("%d\n",ans);
     return ;
 }