POJ 3660 Cow Contest Floyd，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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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 54 34 23 21 22 5

Sample Output

2

 题意  有n头牛  有m对牛有关系 第一头牛打败第二头牛， 求能确定几头牛的排名

    用floyed求传递闭包。如果一个牛和其余的牛关系都是确定的，那么这个牛的排名就是确定的了。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int M=110;int n,m;int dp[M][M];void floyd(){    for(int k=1;k<=n;k++)        for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)        dp[i][j]=dp[i][j]||(dp[i][k]&&dp[k][j]);//传递闭包,记录i和j的关系（直接或间接）是否确定}int main(){    int a,b;    while(~scanf("%d%d",&n,&m))    {        memset(dp,0,sizeof(dp));        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            dp[a][b]=1;        }        floyd();        int ans=0;        for(int i=1;i<=n;i++)        {            int sum=0;            for(int j=1;j<=n;j++)                if(dp[i][j]||dp[j][i])                sum++;            if(sum==n-1)            ans++;        }        printf("%d\n",ans);    }    return 0;}