POJ 3660 Cow Contest

最新推荐文章于 2021-08-06 22:28:19 发布

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本文介绍了一种解决牛群间编程竞赛排名问题的方法。通过分析比赛结果，利用 Floyd 算法确定每头牛之间的胜负关系，最终找出能够确切排名的牛的数量。

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

题意：有许多牛，两头牛之间进行较量，然后我们知道几组较量过后的胜利者，求有几只牛的排名可以确定。

思路：通过floyd确定两只牛之间的关系，如果一只牛输赢总场数为n-1，那么它的排名就是可以确定的。

代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int a,b,ans=0,e[120][120]={0};
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);//e[a][b]表示a能战胜b
            e[a][b]=1;
        }
        for(int k=1; k<=n; k++)
        {
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    if(e[i][k]!=0&&e[k][j]!=0)
                        e[i][j]=1;
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            int sum=0;
            for(int j=1;j<=n;j++)
                sum+=e[i][j]+e[j][i];//i的胜场数+败场数
            if(sum==n-1)
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}