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
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 1<<29
int map[310][310],a[10005],N,M;
int main(){
int n,i,j,k;
while(scanf("%d%d",&N,&M)==2){
for(i=1;i<=N;i++)
for(j=1;j<=N;j++)
if(i==j) map[i][j]=0;
else map[i][j]=map[j][i]=INF;
while(M--){
scanf("%d",&k);
for(i=1;i<=k;i++){
scanf("%d",&a[i]);
for(j=1;j<i;j++)
map[a[i]][a[j]]=map[a[j]][a[i]]=1;
}
}
for(i=1;i<=N;i++)
for(j=1;j<=N;j++)
for(k=1;k<=N;k++)
map[j][k]= min(map[j][k],map[i][k]+map[j][i]);
int ans=INF;
for(i=1;i<=N;i++){
int sum=0;
for(j=1;j<=N;j++)
if(i!=j)
sum+=map[i][j];
ans=min(ans,sum);
}
printf("%d\n",ans*100/(N-1));
}
return 0;
}