POJ2139--Six Degrees of Cowvin Bacon(最简单Floyd)
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
* 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
Sample Input
4 2
3 1 2 3
2 3 4
Sample Output
100
Hint
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std; int way[][]; int main(){
int n,m;
cin>>n>>m;
int temp[];
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
way[i][j]=INT_MAX;
if(j==i)
way[i][i]=;
}
}
for(int i=;i<m;i++){
int num;
cin>>num;
for(int j=;j<num;j++){
cin>>temp[j];
} for(int j=;j<num;j++){
for(int w=j+;w<num;w++){
way[temp[j]][temp[w]]=way[temp[w]][temp[j]]=;
}
}
} for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
for(int w=;w<=n;w++){
if(way[j][i]!=INT_MAX&&way[i][w]!=INT_MAX)
way[j][w]=min(way[j][w],way[j][i]+way[i][w]);
}
}
}
int ans=INT_MAX;
for(int i=;i<=n;i++){
int t=;
for(int j=;j<=n;j++){
t+=way[i][j];
}
ans=min(ans,t);
}
cout<<ans*/(n-);
return ;
}
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