POJ - 2139.Six Degrees of Cowvin Bacon

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#最短路 #POJ - 2139.Six Degrees of Cowv

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本文介绍了一种基于“六度分离”理论的游戏变体，适用于一群参与电影制作的牛。通过构建图模型并利用Floyd算法计算任意两头牛之间的最短路径，最终找出那头牛具有最小的平均度数分离。

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7027		Accepted: 3260

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

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 .]

题解：如果两头牛在同一部电影中出现过，那么这两头牛的度就为1，如果这两头牛a，b没有在同一部电影中出现过，但a，b分别与c在同一部电影中出现过，那么a，b的度为2，即a，b本来不连通map[a][b]=0，加上中介c就联通了map[a][b]=map[a][c]+map[c][b]=2，只要找到这个关系就可以得出两个牛之间的度即为两个牛之间的最短路。给出m部电影，每一部给出牛的个数，和牛的编号。问那一头到其他每头牛的度数平均值最小，输出最小平均值乘100

思路：到所有牛的度数的平均值最小，也就是到所有牛的度数总和最小。那么就是找这头牛到其他每头牛的最小度，也就是最短路径，相加再除以（n-1）就是最小平均值。对于如何确定这头牛，我们可以用Floyd进行遍历，最后记录最下平均值即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
int N,M;
int map[310][310];
void MAP(){//初始化很重要
	for(int i=0;i<310;i++){
		for(int j=0;j<310;j++){
	       map[i][j]=INF;
		}
	}
}
void getmap(){//获取数据，构建地图
    int d[310];
	for(int i=0;i<M;i++){
		int t;
		cin>>t;
		for(int j=0;j<t;j++){
			cin>>d[j];
		}
		for(int j=0;j<t;j++){
			for(int k=0;k<j;k++){
				map[d[j]][d[k]]=1;
				map[d[k]][d[j]]=1;
			}
		}	
	}
}
void floyd(){//floyd算法
	  for(int k=1;k<=N;k++){
   	    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]);
          }
       }
   }
}
int main(){
   while(cin>>N&&N){
   cin>>M;
   MAP();
   getmap();
   floyd();
   int sum,ans=INF;
   for(int i=1;i<=N;i++){
           sum=0;
   	       for(int j=1;j<=N;j++){
   	 	     if(i!=j)
   	 	      sum=sum+map[i][j];
   	         }
   	       ans=min(ans,sum);
   }
   printf("%d\n",int(ans*1.0/(N-1)*100));
  }
}