POJ 3660 传递闭包 floyd

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最新推荐文章于 2024-02-16 23:33:59 发布

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分类专栏： dp 动态规划最短路径

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博客围绕奶牛编程竞赛排名问题展开，有n头奶牛参赛，给出m个两头牛竞赛结果（A战胜B），要确定能明确排名的奶牛数量。思路是若一头牛和其他n - 1头牛关系都明确，其排名可定，可用求传递闭包算法或弗洛伊德算法求解。

http://poj.org/problem?id=3660

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

题目大意：有n头奶牛，给出m个关系，每个关系给出两头牛的编号A和B表示A战胜了B，问能确定排名的牛有多少头。

思路：如果1头牛和其他n-1头牛的关系都知道的话，那么这头牛的排名就可以确定了。因此我们可以用求传递闭包的算法来做。(用弗洛伊德也可以)

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#define INF 0x3f3f3f3f
using namespace std ;

int dp[105][105];

int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
        dp[i][i]=0;
    int t1,t2;
    for(int i=0;i<m;i++)
    {
        scanf("%d %d",&t1,&t2);
        dp[t1][t2]=1;
    }
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                dp[i][j]=dp[i][j]|(dp[i][k]&&dp[k][j]);
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        int sum=0;
        for(int j=1;j<=n;j++)
        {
            if(dp[i][j]||dp[j][i])
                ++sum;
        }
        if(sum==n-1)
            ++ans;
    }
    printf("%d\n",ans);
    return 0;
}