POJ3667（传递闭包）

Cow Contest

时间限制： 1000 ms  |  内存限制： 65535 KB

难度： 4

描述

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.

输入

* 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

There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.

输出

For every case:
* Line 1: A single integer representing the number of cows whose ranks can be determined

样例输入

5 5
4 3
4 2
3 2
1 2
2 5
0 0

样例输出

2

题目大意：有n头牛比赛，m种比赛结果，最后问你一共有多少头牛的排名被确定了，其中如果a战胜b，b战胜c，则也可以说a战胜c，即可以传递胜负。求能确定排名的牛的数目。

关键在于传递胜负。 想到了这一点就不难想到可以完成传递闭包的Floyd算法。

只要满足大于 当前节点 的牛的数量 加上 小于当前节点牛的数量等于 n - 1， 那么当前节点的排名（rank）就可以确定了

上面的条件只要相通，就非常容易编程了。然而笔者并没有想到。。

【CODE】

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#define INF 99999999
using namespace std;
int map[111][111];
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF && (n||m)){
        int a,b;
        memset(map,0,sizeof(map));
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            map[a-1][b-1] = 1;
        }
        for(int k=0;k<n;k++) //传递闭包
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    if(map[i][k] && map[k][j]) //传递
                        map[i][j] = 1;
        int ans =0;
        for(int i=0;i<n;i++){
            int l = 0; // 出度入度的和，也就是大于他，小于他的节点的总数
            for(int j=0;j<n;j++){
                if(map[i][j] || map[j][i])
                    l++;
            }
            if(l == n-1)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}