# POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest（图论，传递闭包）

### 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

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

5 5
4 3
4 2
3 2
1 2
2 5

2

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxN=101;
const int inf=2147483647;

int n,m;
int M[maxN][maxN];

int main()
{
while (cin>>n>>m)
{
memset(M,0,sizeof(M));
for (int i=1;i<=m;i++)
{
int u,v;
cin>>u>>v;
M[u][v]=1;
}
for (int k=1;k<=n;k++)//传递闭包
for (int i=1;i<=n;i++)
for (int j=1;j<=n;j++)
M[i][j]=M[i][j] || ((M[i][k])&&(M[k][j]));
int Ans=0;
for (int i=1;i<=n;i++)
{
int cnt=0;
for (int j=1;j<=n;j++)
if ((i!=j)&&((M[i][j])||(M[j][i])))//M[i][j]就是i打赢j，M[j][i]表示i被j打败，因为不管被打败还是打赢，都是确定了i与j的关系，所以都要统计
cnt++;
if (cnt==n-1)
Ans++;
}
cout<<Ans<<endl;
}
return 0;
}

posted @ 2017-07-24 10:41  SYCstudio  阅读(...)  评论(...编辑  收藏