POJ2139 Six Degrees of Cowvin Bacon

                                                                           Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1891		Accepted: 886

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

* 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

100

Hint

[Cow 3 has worked with all the other cows and thus has degrees of separation: 1, 1, and 1 -- a mean of 1.00 .]

Source

USACO 2003 March Orange

 

 

 思路：本题题意可以变换的理解为如果N个点在一个集合中，则这些点之间的距离为1。然后由此建立一个无向图。在这N个点中，每一个点与其他的所有点都有一个连接的路径长度，将这些长度都加起来，然后除以N-1，就求出了平均长度。题目所求为这些平均长度中的最小值，然后将最小值乘以100输出。

         本题可以先建立邻接矩阵，然后再使用FLOYD就可以求出了。

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>


#define MAXINT 99999999

using namespace std;

int main(int argc, char *argv[])
{
    
    int n,m;
    scanf("%d%d",&n,&m);
    
    int data[304][304];
    
    int array[304];
    
    
    int i,j,k;
    
    
    for(i=1;i<=n;i++)
    for(j=1;j<=n;j++)
    {data[i][j]=MAXINT;}
    
    
    
    
    for(i=1;i<=m;i++)
    {
                     int num;
                     scanf("%d",&num);
                     
                     
                     
                     for(j=1;j<=num;j++)
                     {
                                        
                                        scanf("%d",&array[j]);
                                        
                                        
                      
                                        
                     }
                     
                     
                     for(j=1;j<=num;j++)
                     for(k=j+1;k<=num;k++)
                     data[array[j]][array[k]]=data[array[k]][array[j]]=1;
                     
                     
                     
    }
    
    
    for(k=1;k<=n;k++)
    for(i=1;i<=n;i++)
    for(j=1;j<=n;j++)
    {
                     if(data[i][j]>data[i][k]+data[k][j])
                     data[i][j]=data[i][k]+data[k][j];
    }
    
    
    long long sum=0;
    
    long long minsum=MAXINT;
    
    for(i=1;i<=n;i++)
    {sum=0;
    for(j=1;j<=n;j++)
    {
                     if(i==j)
                     continue;
                     sum+=data[i][j];
    }
    
    if(sum<minsum)
    minsum=sum;
    }
    
    
    
    
    
    
    printf("%d\n",(minsum*100/(n-1)));

                     


                    
                    
                    
                    
                    
    
    
    
    
    
   // system("PAUSE");
    return EXIT_SUCCESS;
}