POJ 2139 - Six Degrees of Cowvin Bacon（floyd）

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2228		Accepted: 1039

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 .]

============================

一起工作的每只之间都有一条距离为1的边，floyd即可

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int INF=0x3f3f3f3f;
int n,m,num;
int vis[333][333],a[333];

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

int main()
{
memset(vis,INF,sizeof(vis));
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
vis[i][i]=0;
for(int k=0;k<m;k++)
{
scanf("%d",&num);
for(int i=0;i<num;i++)
scanf("%d",&a[i]);
for(int i=0;i<num;i++)
for(int j=i+1;j<num;j++)
vis[a[i]][a[j]]=vis[a[j]][a[i]]=1;
}
floyd();
double ans=1e9;
for(int i=1;i<=n;i++)
{
double sum=0;
for(int j=1;j<=n;j++)
{
sum+=vis[i][j];
}
sum/=n-1;
ans=min(ans,sum);
//cout<<"ans="<<ans<<endl;
}
printf("%d\n",(int)(ans*100));
return 0;
}