poj 3660 传递闭包问题

CowContest竞赛解析

最新推荐文章于 2020-05-06 15:48:38 发布

转载最新推荐文章于 2020-05-06 15:48:38 发布 · 159 阅读

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原文链接：http://www.cnblogs.com/SparkPhoneix/p/9444152.html

本文详细解析了CowContest竞赛的算法实现，介绍了如何通过传递闭包算法确定参赛牛的技能排名，提供了一段C++代码示例，展示了如何利用矩阵进行技能比较，最终确定可以精确排名的牛的数量。

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15540		Accepted: 8657

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

题目意思：有一群牛，厉害的要排在前面，输入a，b表示a比b要厉害，问有多少不能确定排名的牛，一道简单的传递闭包的题。

传递闭包大概就是。。 
先将一张图用一个矩阵表示出来，矩阵中的a[i][j]=1表示i~j有一条直接相连的边。
这样就得到一个0/1矩阵。传递闭包算法的目的就是根据以上的初始矩阵，探索出最终的矩阵，
表示根据初始的直接连接关系，从初始矩阵扩展出一个包括间接连接关系的最终矩阵。这个最终矩阵就是传递闭包矩阵
代码如下：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
#define M 20005
#define N 105
int map[N][N];
void fold(int sum)
{
    for(int i=1;i<=sum;i++)
        for(int j=1;j<=sum;j++)
        for(int k=1;k<=sum;k++)
        map[j][k]=map[j][k]||(map[j][i]&&map[i][k]);
}
int main()
{
    int n,m;
    cin>>n>>m;
    int a,b;
    memset(map,0,sizeof(map));
    for(int i=0;i<m;i++)
    {
        cin>>a>>b;
        map[a][b]=1;
    }
    fold(n);
    int res=0;
    for(int i=1;i<=n;i++)
       {
           int ans=0;
        for(int j=1;j<=n;j++)
    {
        if(i==j)continue;
        if(map[i][j]||map[j][i])ans++;
    }
    if(ans==n-1)res++;
       }
    cout<<res<<endl;

}

转载于:https://www.cnblogs.com/SparkPhoneix/p/9444152.html