POJ 2139 Six Degrees of Cowvin Bacon(Floyd)

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本文介绍了一种算法，用于计算一群参与共同电影项目的牛之间的平均分离度。通过分析每头牛与其他所有牛之间的最短距离之和来找出具有最小平均距离的牛。

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The cows have been making movies lately, so they are ready to play a variant of the famous game "Six Degrees of Kevin Bacon".

The game works like this: each cow is considered to be zero degrees of separation (degrees) away from herself. If two distinct cows have been in a movie together, each is considered to be one 'degree' away from the other. If a two cows have never worked together but have both worked with a third cow, they are considered to be two 'degrees' away from each other (counted as: one degree to the cow they've worked with and one more to the other cow). This scales to the general case.

The N (2 <= N <= 300) cows are interested in figuring out which cow has the smallest average degree of separation from all the other cows. excluding herself of course. The cows have made M (1 <= M <= 10000) movies and it is guaranteed that some relationship path exists between every pair of cows.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each input line contains a set of two or more space-separated integers that describes the cows appearing in a single movie. The first integer is the number of cows participating in the described movie, (e.g., Mi); the subsequent Mi integers tell which cows were.

Output

* Line 1: A single integer that is 100 times the shortest mean degree of separation of any of the cows.

Sample Input

4 2
3 1 2 3
2 3 4

Sample Output

题目大意是这么多个牛，在一起拍过电影距离为1，那么到其他牛最小的距离和的平均值最小的是多少？且输出这个值乘以100.

代码如下：

#include <iostream>
#include <cstdio>
using namespace std;

int n,m;
int ma[305][305];
const int INF=99999999;

void init()
{
    for(int i=1;i<=304;i++)
        for(int j = 1; j <= 304;j++)
        if( i != j )ma[i][j] = INF;
}

int main()
{
    int temp[301];
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        init();
        for(int i=0;i<m;i++)
        {
            int t;
            scanf("%d",&t);

            for(int j=0;j<t;j++)
                scanf("%d",&temp[j]);

            for(int j=0;j<t;j++)
                for(int k=0;k<t;k++)
                if(j!=k)ma[temp[j]][temp[k]]=1;
        }
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
            for(int j = 1;j <= n;j++)
            ma[i][j] = min(ma[i][j],ma[i][k] + ma[k][j]);
        int ans=INF;
        for(int i = 1;i <= n;i++)
        {
            int sum=0;
            for(int j = 1;j <= n;j++)
                sum += ma[i][j];
            ans=min(sum,ans);
        }
        cout<<ans*100/(n-1)<<endl;

    }
}