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[][];
 int n, m;
 int main()
 {
     int n, m;
     cin >> n >> m;
     int i, j;
     int a[];
     for (i = ; i <= n; i++)
     {
         for (j = ; j <= n; j++)
         {
             if (i == j) e[i][j] = ;
             else e[i][j] = inf;
         }
     }
     for (i = ; i <= m; i++)
     {
         int k,p;
         cin >> k;
         for (j = ; j <= k; j++)
         {
             cin >> a[j];
         }
         for (j = ; j <= k; j++)
         {
             for (p = j+; p <= k; p++)
             {
                 e[a[j]][a[p]] = ;
                 e[a[p]][a[j]] = ;
             }
         }
     }
     int k;
     for (k = ; k <= n; k++)
     {
         for (i = ; i <= n; i++)
         {
             for (j = ; 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 = ;
     for (i = ; i <= n; i++)
     {
         sum = ;
         for (j = ; j <= n; j++)
         {
             if (e[i][j] != inf)
             {
                 sum += e[i][j];
             }
         }
         if (sum < ans) ans = sum;
     }
     ans= ans*/ (n-) ;
     cout <<ans<< endl;
 }