本文主要是介绍POJ - 3660，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！
Cow Contest
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
最短路径变形 通过输赢的关系进行松弛操作把所有可以知道的每头牛之间的关系列出来然后判断这头牛是否和其他n-1头牛有关系如果有就可以判断他的等级。
#include <iostream>
#include <cmath>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
int a[105][105],c,n,mid;
int main()
{int k,u,m;while(cin>>n>>c){memset(a,0,sizeof(a));for(int i=0;i<c;i++){cin>>u>>m;a[u][m]=1;}for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){a[i][j]=a[i][j]||(a[i][k]&&a[k][j]);}}}int s=0;for(int i=1;i<=n;i++){int sum=0;for(int j=1;j<=n;j++){if(a[i][j]||a[j][i])sum++;}if(sum==n-1)s++;}cout<<s<<endl;}
}
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