poj 2139 Six Degrees of Cowvin Bacon

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5766		Accepted: 2712

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

/*
* @Author: Lyucheng
* @Date:   2017-07-21 10:09:45
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-07-21 10:32:45
*/
/*
 题意：有n头奶牛，如果奶牛在同一部电影中工作过，那么这些奶牛就是有关系的，关系距离为1
    当然也可以通过其他奶牛产生间接的关系，问你那头奶牛的与其他奶牛的平均关系距离最小

 思路：建边然后最短路
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define MAXN 305
#define INF 10000000
using namespace std;

int n,m;
int t;
int pos[MAXN];
int dp[MAXN][MAXN];

void floyd(){
    for(int i=1;i<=n;i++){
        for(int k=1;k<=n;k++){
            if(k==i) continue;
            for(int j=1;j<=n;j++){
                if(j==i) continue;
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
            }
        }
    }
}

void init(){
    for(int i=0;i<=n;i++){
        for(int j=0;j<=n;j++){
            dp[i][j]=INF;
        }
        dp[i][i]=0;
    }
}

int main(){ 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        for(int i=0;i<m;i++){//建边
            scanf("%d",&t);
            for(int j=0;j<t;j++){
                scanf("%d",&pos[j]);
            }
            for(int j=0;j<t;j++){
                for(int k=j+1;k<t;k++){
                    dp[pos[j]][pos[k]]=100;
                    dp[pos[k]][pos[j]]=100;
                }
            }
        }
        floyd();
        int _min=INF;
        for(int i=1;i<=n;i++){
            int cur=0;
            for(int j=1;j<=n;j++){
                if(j==i) continue;
                cur+=dp[i][j];
            }
            _min=min(_min,cur/(n-1));
        }
        printf("%d
",_min);
    }
    return 0;
}