poj--2139

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

题意：奶牛们要拍电影，如果两个奶牛同时拍一部那么他们的距离为1，如果两个奶牛同时和第三头奶牛拍一部电影，那么他们的距离为2，求他们最短距离乘上100的最小平均值
　　
这题是相当于求最短路径，forld最小环从自身出发，最后到达自身的最短路径，
Flord 最小环问题相当于一个动态规划的算法，如果a不能到达b那么可以考虑引入一个k，经过k到达b并更新一下a，b的距离
所以这个算法的核心代码是
　　for(int k=1;k<=n;k++)
　　　　for(int i=1;i<=n;i++)
　　　　　　for(int j=1;j<=n;j++)
　　　　　　　　map[i][j]=min(map[i][j],map[i][k]+map[k][j])
最近做题一直忘记预处理，导致程序不对。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int Max_n=;
const int INF=;
int map[Max_n][Max_n];
int node[Max_n];
int n,m;
void prepare()
{
    for(int i=;i<=n;i++)
        for(int j=;j<=n;j++)
        {
            if(i!=j)    map[i][j]=map[j][i]=INF;
        }
}
int main()
{
    while(cin>>n>>m)
    {
        prepare();
        int x;
        while(m--)
        {
            cin>>x;
            for(int i=;i<=x;i++)
            {
                cin>>node[i];
            }
            for(int i=;i<=x;i++)
                for(int j=i+;j<=x;j++)
                map[node[i]][node[j]]=map[node[j]][node[i]]=;
            for(int k=;k<=n;k++)
                for(int i=;i<=n;i++)
                    for(int j=;j<=n;j++)
                map[i][j]=min(map[i][j],map[i][k]+map[k][j]);
        }
        int ans=INF;
            for(int i=;i<=n;i++)
            {
                int maxn=;
                for(int j=;j<=n;j++)
                {
                    maxn+=map[i][j];
                }
                ans=min(ans,maxn*/(n-));
            }
        printf("%d\n",ans);
    }
    return ;
}