POJ2139--Six Degrees of Cowvin Bacon(最简单Floyd)

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 .] 

 

 

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

int way[305][305];

int main(){
    int n,m;
    cin>>n>>m;
    int temp[305];
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            way[i][j]=INT_MAX;
            if(j==i)
                way[i][i]=0;
        }
    }
    for(int i=0;i<m;i++){
        int num;
        cin>>num;
        for(int j=0;j<num;j++){
            cin>>temp[j];
        }
        
        for(int j=0;j<num;j++){
            for(int w=j+1;w<num;w++){
                way[temp[j]][temp[w]]=way[temp[w]][temp[j]]=1;
            }
        }
    }
    
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int w=1;w<=n;w++){
                if(way[j][i]!=INT_MAX&&way[i][w]!=INT_MAX)
                    way[j][w]=min(way[j][w],way[j][i]+way[i][w]);
            }
        }
    }
    int ans=INT_MAX;
    for(int i=1;i<=n;i++){
        int t=0;
        for(int j=0;j<=n;j++){
            t+=way[i][j];
        }
        ans=min(ans,t);
    }
    cout<<ans*100/(n-1); 
    return 0;
}