POJ 2139 Six Degrees of Cowvin Bacon (floyd）

Six Degrees of Cowvin Bacon

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 1

Problem Description

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 <br> <br>* 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. <br>

 

Output

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

 

Sample Input

4 2 3 1 2 3 2 3 4

 

Sample Output

100

 

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#define inf 0x3f3f3f3f
using namespace std;
int e[305][305];
int n, m;
int main()
{
    int n, m;
    cin >> n >> m;
    int i, j;
    int a[305];
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (i == j) e[i][j] = 0;
            else e[i][j] = inf;
        }
    }
    for (i = 1; i <= m; i++)
    {
        int k,p;
        cin >> k;
        for (j = 1; j <= k; j++)
        {
            cin >> a[j];
        }
        for (j = 1; j <= k; j++)
        {
            for (p = j+1; p <= k; p++)
            {
                e[a[j]][a[p]] = 1;
                e[a[p]][a[j]] = 1;
            }
        }
    }
    int k;
    for (k = 1; k <= n; k++)
    {
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
            {
                if (e[i][j] > e[i][k] + e[k][j])
                    e[i][j] = e[i][k] + e[k][j];
            }
        }
    }
    int ans = inf;
    int sum = 0;
    for (i = 1; i <= n; i++)
    {
        sum = 0;
        for (j = 1; j <= n; j++)
        {
            if (e[i][j] != inf)
            {
                sum += e[i][j];
            }
        }
        if (sum < ans) ans = sum;
    }
    ans= ans*100/ (n-1) ;
    cout <<ans<< endl;
}