本文主要是介绍HD Problem A ，POJ 3660 Cow Contest.(Floyd，传递闭包）.，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！
Problem A
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 28 Accepted Submission(s) : 10
Problem 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
<p>* Line 1: Two space-separated integers: <i>N</i> and <i>M</i><br>* Lines 2..<i>M</i>+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, <i>A</i>, is the winner) of a single round of competition: <i>A</i> and <i>B</i></p>
Output
<p>* Line 1: A single integer representing the number of cows whose ranks can be determined<br> </p>
Sample Input
5 54 34 23 21 22 5
Sample Output
2
题意：给定几对牛之间的关系，求出有几个牛的关系是唯一确定的：floyed算法的变形应用 ：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int f[101][101];
int main()
{int n,m,win,lose;scanf("%d%d",&n,&m);memset(f,0,sizeof(f));for(int i=1;i<=m;i++){scanf("%d%d",&win,&lose);f[win][lose]=1;}for(int k=1;k<=n;k++)//floyed算法变形for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)f[i][j]=f[i][j]||(f[i][k]&&f[k][j]);//确定i，j两点之间是否相连int sum=0;for(int i=1;i<=n;i++){int t=0;for(int j=1;j<=n;j++){if(f[i][j]||f[j][i]) t++;//}if(t==n-1) sum++;}printf("%d\n",sum);return 0;
}
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