POJ3660:Cow Contest（Floyd传递闭包）

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16941		Accepted: 9447

题目链接：http://poj.org/problem?id=3660

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

Sample Input:

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

Sample Output:

2

题意：

有n个人，m场比赛，然后给出m场比赛的胜负关系，问有多少只牛能确定它们自己的名次。

题解：

这个题有点像拓扑排序，但是只用拓扑序并不能保证结果的正确性。

其实解这个题我们只需要发现这样一个关系就好了，若一只牛的名次能够被确定，那么它赢它的牛和它赢的牛个数之和为n-1。

利用这个关系，我们floyd传递闭包预处理一下，然后判断一下数量关系就好了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const int N = , M = ;
int n,m;
int mp[N][N];
int main(){
    scanf("%d%d",&n,&m);
    for(int i=;i<=m;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        mp[u][v]=;
    }
    for(int k=;k<=n;k++){
        for(int i=;i<=n;i++){
            for(int j=;j<=n;j++){
                mp[i][j]=(mp[i][j]|(mp[i][k]&mp[k][j]));
            }
        }
    }
    int ans=;
    for(int i=;i<=n;i++){
        int win=,lose=;
        for(int j=;j<=n;j++){
            if(mp[i][j]) win++;
            if(mp[j][i]) lose++;
        }
        if(win+lose==n-) ans++;
    }
    cout<<ans;
    return ;
}