POJ 3660 Cow Contest (Floyd 传递闭包判连通)

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本文介绍了一道关于编程竞赛中牛的排名问题的算法题。通过建立胜负关系图，并运用Floyd算法，最终确定每头牛的排名。适用于了解图论及排名算法的读者。

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7311		Accepted: 4045

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

题目链接：http://poj.org/problem?id=3660

题目大意：n个人m个关系，a b表示a胜b，问有多少人的名次确定

题目分析：先根据胜负关系建图，map[a][b]表示a名次在b之前，名次具有传递性，也就是如果a胜b，b胜c则a的名次一定在c之前，所以建图完跑一遍Floyd，然后根据点的度数，若度数为n－1，说明其他人与他的关系都确定，则他的名次就可以确定了

#include <cstdio>
#include <cstring>
int m, n, map[105][105], d[105];

void Floyd()
{
    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])
                    map[i][j] = 1;
}

void cal()
{
    for(int i = 1; i<= n; i++)
    {
        for(int j = 1; j<= n; j++)
        {
            if(map[i][j])
            {
                d[i] ++;
                d[j] ++;
            }
        }
    }
}

int main()
{
    int a, b, ans = 0;
    memset(map, 0, sizeof(map));
    memset(d, 0, sizeof(d));
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++)
    {
        scanf("%d %d", &a, &b);
        map[a][b] = 1;
    }
    Floyd();
    cal();
    for(int i = 1; i <= n; i++)
        if(d[i] == n - 1)
            ans ++;
    printf("%d\n", ans);
}