Cow Contest POJ - 3660
题目：
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：
输入奶牛的数量N，奶牛间PK的回合数M，M个回合的结果（前者胜，后者负），问最终可以确定几头奶牛的排名。
思路：
N<=100，考虑floyed。
利用传递性质，判断图中各点到剩余所有点的是否联通（存在单项联通即可），输出这样的点的数量。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=105;
int N,M,a[maxn][maxn],ans=0;int main()
{cin>>N>>M;memset(a,0,sizeof(a));for(int i=0;i<M;i++){int u,v;scanf("%d%d",&u,&v);a[u][v]=1;}for(int k=1;k<=N;k++)for(int i=1;i<=N;i++)for(int j=1;j<=N;j++)if(a[i][k]&&a[k][j])a[i][j]=1;//如果i打败k，k打败j，则i打败j-----传递闭包for(int i=1;i<=N;i++){int tmp=0;//第一层循环内，每个点需要重新置零for(int j=1;j<=N;j++){if(a[i][j]||a[j][i])//i与j之间的胜负关系确定（联通）tmp++;}if(tmp==N-1) ans++;//与剩余N-1个点的关系都确定（都存在单向联通）}cout<<ans<<endl;return 0;
}