POJ3660 Cow Contest(Floyd,传递闭包)

2017年3月3日 1条评论 652次阅读 0人点赞

题目：

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10767		Accepted: 5996

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

Sample Output

Source

USACO 2008 January Silver

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思路：

有n头牛，进行了m次比赛，然后有m组比赛信息，第一个为胜利者。问最后有几头牛的排名可以确定，利用folyd吧间接的关系变成直接关系，用传递闭包,对每给的一个胜负关系连一条边，最后跑一次Floyd，然后判断一头牛所确定的关系是否是n-1次，若是，则这头牛的排名可以确定

代码：

#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <cmath>  
#include <string>  
#include <iostream>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <algorithm>  
#define mem(a,b) memset(a,b,sizeof(a))  
#define N 100+20  
#define M 100000+20  
#define inf 0x3f3f3f3f  
using namespace std;  
int n,m;  
int map[N][N];  
int main()  
{  
    int a,b;  
    while(~scanf("%d%d",&n,&m))  
    {  
        mem(map,0);  
        for(int i=1; i<=m; i++)  
        {  
            scanf("%d%d",&a,&b);  
            map[a][b]=1;  
        }  
        int sum=0;  
        for(int k=1; k<=n; k++)  
            for(int i=1; i<=n; i++)  
                for(int j=1; j<=n; j++)  
                    if(map[i][k]&&map[k][j])//间接变成直接  
                        map[i][j]=1;  
                  
        for(int i=1; i<=n; i++)  
        {  
            int k=0;  
            for(int j=1; j<=n; j++)  
                k+=map[i][j]+map[j][i];  
            if(n-1==k)//如果一头牛和其他所有牛的关系确定了的话，它的排名也就确定了。    
                sum++;  
        }  
        printf("%d\n",sum);  
    }  
    return 0;  
}