Six Degrees of Cowvin Bacon （最短路）

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 .] 

思路：最短路问题。

在建图的时候要注意将对角线一列设为0，其余在初始状态下设为INF无穷大

然后将在一个集合的元素之间的距离设为1（这里的图为对称矩阵！！）

之后再用Floyd算法更新间接有关系的坐标Map[i][j]=min(Map[i][j],Map[i][k]+Map[k][j]);

最后输出Min*100/(n-1)（这里注意，一定要先乘100，再去取平均，int的数不能整除的话会舍去小数点后面的数，导致结果扩大100倍后误差较大！！！）

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 99999999
using namespace std;
int main()
{
int n,m,t,a[305];
int Map[305][305];
scanf("%d %d",&n,&m);
for(int i=0;i<=n;i++)
{
Map[i][i]=0;
for(int j=i+1;j<=n;j++)
Map[i][j]=Map[j][i]=INF;
}
for(int i=1;i<=m;i++)
{
scanf("%d",&t);
for(int j=1;j<=t;j++)
scanf("%d",&a[j]);
for(int x=1;x<=t;x++)
for(int y=x+1;y<=t;y++)
Map[a[x]][a[y]]=Map[a[y]][a[x]]=1;
}
for(int k=1;k<=n;k++)//Floyd算法
{
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]);
}
}
}
/*for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++)
printf("%10d",Map[i][j]);
printf("\n");
}*/
int Min=INF,temp;
for(int i=1;i<=n;i++)
{
temp=0;
for(int j=1;j<=n;j++)
temp+=Map[i][j];
if(temp<Min)
Min=temp;
}
printf("%d\n",Min*100/(n-1));//重要点，先乘后除，防止误差过大
return 0;
}