C/B

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 23 1 2 32 3 4

Sample Output

100

牛与牛

#include <iostream>#include<cstring>using namespace std;const double INF=0x3f3f3f;const int maxn=305;int dis[maxn];int vis[maxn];int map[maxn][maxn];int m,n;int a[maxn];void dii(int x){  for(int i=1;i<=n;i++)    dis[i]=INF;  for(int i=1;i<=n;i++)    dis[i]=map[x][i];   vis[x]=1;  for(int i=2;i<=n;i++)  {      int next=0,min=INF;      for(int j=1;j<=n;j++)      {          if(!vis[j]&&min>dis[j])          {next=j;          min=dis[j];          }      }      vis[next]=1;      for(int j=1;j<=n;j++)      {          if(!vis[j]&&dis[j]>map[next][j]+dis[next])            dis[j]=map[next][j]+dis[next];      }  }}int main(){    cin>>n>>m;    for(int i=1;i<=n;i++)        for(int j=1;j<=n;j++)         map[i][j]=INF;    while(m--)    {        int num;        cin>>num;        for(int i=0;i<num;i++)           cin>>a[i];        for(int i=0;i<num;i++)            for(int j=i+1;j<num;j++)             {   if(a[i]!=a[j])                 map[a[i]][a[j]]=map[a[j]][a[i]]=1;             }    }    double ans=INF;    for(int i=1;i<=n;i++)    {        double sum=0;        memset(vis,0,sizeof(vis));        dii(i);        for(int j=1;j<=n;++j)//记录到其他牛的度数之和{if(i!=j)sum+=dis[j];}ans=min(ans,sum*1.0/(n-1));    }cout<<((int)(ans*100))<<endl;    return 0;}