POJ3660——Cow Contest(Floyd+传递闭包)

Cow Contest

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

题目大意：

　　　　给一些牛的排名关系，问有多少牛的排名确定。

解题思路：

　　　　使用Floyd算法来判断传递闭包。

　　　　首先通过输入信息建立邻接矩阵，再使用Floyd求出最短路径Edge[i][j]。

　　　　这时，相对于牛i若Edge[i][j]存在，则说明i肯定打不过j。若Edge[j][i]存在则说明i肯定打得过牛j。

　　　　若i肯定能打过的牛和肯定打不过的牛的和等于牛的总和N-1，则牛i的位置确定。

　　　　据说这个东西叫做传递闭包--！

Code：

#include<stdio.h>
#include<string>
#include<iostream>
#define MAXN 300
using namespace std;
int edge[MAXN+10][MAXN+10];
int N,M;
void init()
{
    for (int i=1; i<=N; i++)
        for (int j=1; j<=N; j++)
            edge[i][j]=INT_MAX;
}
void floyd()
{
    int m,i,j;
    for (m=1; m<=N; m++)
        for (i=1; i<=N; i++)
            for (j=1; j<=N; j++)
            {
                if (edge[i][m]!=INT_MAX&&edge[m][j]!=INT_MAX&&edge[i][j]>edge[i][m]+edge[m][j])
                    edge[i][j]=edge[i][m]+edge[m][j];
            }
}
int main()
{
    while (cin>>N>>M)
    {
        init();
        for (int i=1; i<=M; i++)
        {
            int x1,x2;
            scanf("%d %d",&x1,&x2);
            edge[x1][x2]=1;
        }
        floyd();
        int sum=0;
        for (int i=1; i<=N; i++)
        {
            int cnt=0;
            for (int j=1; j<=N; j++)
            {
                if (i!=j&&edge[i][j]!=INT_MAX)
                    cnt++;
                if (i!=j&&edge[j][i]!=INT_MAX)
                    cnt++;
            }
            if (cnt==N-1) sum++;
        }
        printf("%d
",sum);
    }
    return 0;
}