POJ-2139-Six Degrees of Cowvin Bacon

Six Degrees of Cowvin Bacon

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7024		Accepted: 3259

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 .] 

Source

 

题意：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。以此类推，a与b之间有n头媒介牛，那么a，b的度为n+1。 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枚举，最后记录最下平均值即可。

思路来自：http://www.cnblogs.com/demian/p/7388642.html

这个题说的真是不清不楚

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 350
#define inf 0x3f3f3f3f
int map[N][N],n,m,a[N];
void floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(map[i][j]>map[i][k]+map[k][j])
                {
                    map[i][j]=map[i][k]+map[k][j];
                }
            }
        }
    }
}
int main()
{
   /* freopen("a.in","rb",stdin);
    freopen("a.out","wb",stdout);*/
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j) map[i][j]=0;
                else map[i][j]=map[j][i]=inf;
            }
        }
        int k;
        for(int i=0;i<m;i++)
        {
            cin>>k;
            for(int i=1;i<=k;i++)
            {
                cin>>a[i];
            }
            for(int i=1;i<=k;i++)
            {
                for(int j=i+1;j<=k;j++)
                {
                    map[a[i]][a[j]]=map[a[j]][a[i]]=1;
                }
            }
        }
        floyd();
        int minn=inf;
        for(int i=1;i<=n;i++)
        {
            int sum=0;
            for(int j=1;j<=n;j++)
            {
                sum+=map[i][j];
            }
            if(sum<minn) minn=sum;
        }
        cout<<minn*100/(n-1)<<endl;//先乘一百再求平均
    }

    return 0;
}

先乘100再求平均 增加精确度