POJ - 3660 Cow Contest (Warshall算法求传递闭包）

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#算法 #闭包 #poj #Warshall

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本文介绍了一种使用Warshall算法解决编程竞赛中牛的技能排名问题的方法。通过构建牛之间的比赛胜负图，并利用Warshall算法求得传递闭包，从而确定能够精确排名的牛的数量。

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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

Sample Output

根据每头牛之间的关系生成一张图， Warshall算法得到传递闭包。若一头牛与其它每头牛之间关系确定，则这头牛的排名就确定。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define N 110
#define INF 0x3f3f3f3f
using namespace std;
bool Map[N][N];
int n, m;
int main()
{
    while(cin >> n >> m)
    {
        memset(Map, false, sizeof(Map));
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            Map[u][v] = true;
        }
        for(int i = 1; i <= n; i++)// Warshall
        {
            for(int j = 1; j <= n; j++)
            {
                if(Map[j][i])
                {
                    for(int k = 1; k <= n; k++)
                    {
                        Map[j][k] += Map[i][k];
                    }
                }
            }
        }
        int j;
        int cnt = 0;
        for(int i = 1; i <= n; i++)//枚举每头牛与其它牛间的关系， 只要与其中一只不能确定， 则不能确定其排名
        {
            for(j = 1; j <= n; j++)
            {
                if(i == j)
                continue;
                if(Map[i][j] || Map[j][i]);
                else
                break;
            }
            if(j > n)
            cnt++;
        }
        printf("%d\n", cnt);
    }
    return 0;
}