POJ 3660 - Cow Contest(传递闭包+Floyd)

最新推荐文章于 2018-06-29 18:51:00 发布

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本文介绍了一种通过已知胜负关系来确定牛群中每头牛技能排名的算法。使用传递闭包结合Floyd算法，当一头牛与其他所有牛之间的胜负关系明确时，该牛的排名即可确定。

Cow Contest
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 10479 Accepted: 5853
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

题意：
给出n的牛和m对关系（A牛可以战胜B牛），问最后有多少个牛的rank可以确定。

解题思路：
传递闭包+floyd，如果当前的牛和n-1头牛的关系都存在，那么这头牛的rank就知道了。
并且如果a和b有关系，b和c有关系，那么a和c也有关系。

AC代码：

#include<stdio.h>
#include<string.h>
const int maxn = 1e2+5;
bool mp[maxn][maxn];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    memset(mp,0,sizeof(mp));
    while(m--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        mp[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(mp[i][k] && mp[k][j])    mp[i][j] = 1;
    int ans = 0;
    int j;
    for(int i = 1;i <= n;i++)
    {
        for(j = 1;j <= n;j++)
        {
            if(i == j)  continue;
            if(!mp[i][j] && !mp[j][i])  break;
        }
        if(j > n)   ans++;
    }
    printf("%d",ans);
    return 0;
}