POJ-3660 Cow Contest

Problem 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

有一群牛，他们之间两两进行比赛，现在只知道他们中某几个的输赢关系，问能确定几只牛的排名？

解法

这道题要用的传递闭包，只要把他们的传递闭包求出来（这里用到了Folyd-WarShall算法，因为若牛A打败了牛B,牛B又打败了牛C,那么可以知道牛A也能打败牛C,在这里就表示为可达），然后计算其中出度与入度之和为N-1的结点的数目就可以知道那几只牛的排名可以确定了（因为那只牛被某几只牛打败了，又打败了某几只牛，所以。。。）

Solution

#include<iostream>
using namespace std;
#define MAX 100
int G[MAX][MAX];
int WarShall(int G[][MAX], int n) {
    int count = 0;
    for (int k = 0; k < n; k++){
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (G[i][k] == 1 && G[k][j] == 1) {
                    G[i][j] = 1;
                }
            }
        }
    }

    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = 0; j < n; j++) {
            if (G[i][j] == 1||G[j][i]==1) {    //如果打败了谁或被谁打败了
                sum++;                         //也就是求出度与入度之和

            }
        }
        if (sum == n - 1) {
            count++;
        }
    }
    return count;
}
int main() {
    int n, m;
    int u, v;
    int result;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> u >> v;
        G[u-1][v-1] = 1;
    }
    result = WarShall(G, n);
    cout << result;
    system("pause");
}

原文地址：https://www.cnblogs.com/urahyou/p/11406282.html