本文主要是介绍Cow Contest POJ-3660(floyd)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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
题意：有许多牛，两头牛之间进行较量，然后我们知道几组较量过后的胜利者，求有几只牛的排名可以确定。
思路：通过floyd确定两只牛之间的关系，如果一只牛输赢总场数为n-1，那么它的排名就是可以确定的。
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
int e[110][110];
int main()
{int n,m;while(~scanf("%d%d",&n,&m)){memset(e,0,sizeof e);for(int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);e[x][y]=1;}for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)if(e[i][k]&&e[k][j])e[i][j]=1;int ans=0;for(int i=1;i<=n;i++){int sum=0;for(int j=1;j<=n;j++)sum+=e[i][j]+e[j][i];//当一头牛与其他所有牛的关系确定后 它的位置就确定了 if(sum==n-1)//输的次数和赢的次数相加等于n-1 即能能确定其与其他所有牛的关系ans++;}printf("%d\n",ans);}return 0;
}
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