POJ 3660 传递闭包问题

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

题意概述

　　有n头牛比赛，给出m个胜负关系，最后问你一共有多少头牛的排名被确定了。特别的，其中如果a战胜b，b战胜c，则也可以说a战胜c，就是说胜负是可以传递的。求能确定排名的牛的数目。

题解

　　一个牛的排名被确定了，当且仅当它被战胜的次数和它战胜别人的次数之和为n-1。所以这题就转化成了求传递闭包问题，很H2O的啊QAQ

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 105;
int map[M][M];
int n, m;
int main()
{
    cin>>n>>m;
    int a, b;
    for(int i=0; i<m; ++i)
    {
        cin>>a>>b;
        map[a][b] = 1;
    }
    for(int k=1; k<=n; ++k)
        for(int i=1; i<=n; ++i)
            if(map[i][k] == 1)
                for(int j=1; j<=n; ++j)
                    if(map[k][j] == 1)
                        map[i][j] = 1;
    int cnt, ans = 0;
    for(int i=1; i<=n; ++i)
    {
        cnt = 0;
        for(int j=1; j<=n; j++)
            if(map[i][j] == 1 || map[j][i] == 1) cnt++;
        if(cnt == (n-1)) ans++;
    }
    cout<<ans<<endl;
    return 0;
}