POJ 2139

简单枚举+FLOYD

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string.h>
#include <queue>
#include <cmath>
#include <vector>
using namespace std;
const int N=310;
const int inf=(1<<30)-1;

int dist[N][N];
int cow[N];
int n,m;

int main(){
	int c;
	while(scanf("%d%d",&n,&m)!=EOF){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(i==j) dist[i][j]=0;
				else dist[i][j]=inf;
			}
		}
		for(int i=1;i<=m;i++){
			scanf("%d",&c);
			for(int k=1;k<=c;k++)
			scanf("%d",&cow[k]);
			for(int p=1;p<=c;p++){
				for(int q=p+1;q<=c;q++){
					dist[cow[p]][cow[q]]=dist[cow[q]][cow[p]]=1;
				}
			}
		}
		for(int k=1;k<=n;k++){
			for(int i=1;i<=n;i++){
				for(int j=1;j<=n;j++)
				if(dist[i][j]>dist[i][k]+dist[k][j])
				dist[i][j]=dist[i][k]+dist[k][j];
			}
		}
		int ans=inf;
		for(int i=1;i<=n;i++){
			int tmp=0;
			for(int j=1;j<=n;j++)
			tmp+=dist[i][j];
			if(ans>tmp) ans=tmp;
		}
		printf("%d\n",int(ans*100.0/(n-1)));
	}
	return 0;
}