poj_3660_Cow Contest

最新推荐文章于 2023-07-18 14:39:56 发布

米先森

最新推荐文章于 2023-07-18 14:39:56 发布

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分类专栏： ACM刷题文章标签： poj

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本文介绍了一个基于传递闭包原理的算法，用于解决牛之间比赛后的精确排名问题。通过记录每场比赛的结果并应用沃舍尔算法更新传递闭包矩阵，最终确定能够明确排名的牛的数量。

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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

题意：
题意：有n 头牛比赛，m 种比赛结果，最后输出可以确定排名的牛的个数。
比如：牛a战胜b,b战胜c，即a战胜c，……胜负可以传递。
如有三头牛，a战胜c,a战胜b,输出结果为1.
分析：
如果其中一头牛被i头牛打败(e[i][x])，而战胜了j头牛(e[x][j])，则x+y = n-1;则这头牛的排名就确定了；
这个题需要利用传递闭包问题：
此题用来判断比赛中名次排列的。
传递闭包问题：一个有n个顶点的有向图的传递闭包为：邻接矩阵中e[i,j]表示i到j是否直接可达，若直接可达，则e[i,j]记为1，否则记为0；有向图的传递闭包表示从邻接矩阵e出发，求的所有节点间的路径可达情况，该矩阵就为所要求的传递闭包矩阵。

//传递闭包
void Warshll()
{
    int i,j,k;
    for(k=1;k<=n;k++)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
                e[i][j] = e[i][j] || (e[i][k]&&e[k][j]);
        }
    }
}

代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int e[505][505];
int n,m;
//传递闭包
void Warshll()
{
    int i,j,k;
    for(k=1;k<=n;k++)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
                e[i][j] = e[i][j] || (e[i][k]&&e[k][j]);
        }
    }
}
int main()
{
    int i,j;
    int A,B;
    scanf("%d%d",&n,&m);
    memset(e,0,sizeof(e));//初始化为0
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&A,&B);
        e[A][B]=1;//有向图，用1表示从i 到j可以连通。
    }

    Warshll();
    int ans,res=0;
    for(i=1;i<=n;i++)
    {
        ans=0;
        for(j=1;j<=n;j++)
        {
            if(i==j)
                continue;
            else if(e[i][j]==1 || e[j][i]==1)//被打败和战胜的次数，i,j有两种情况，要么被i打败（失败）,要么打败i(战胜);
                ans++;  
        }
        if(ans==n-1)
            res++;
    }
    printf("%d\n",res);


    return 0;
}