Poj 2139 Six Degrees of Cowvin Bacon （floyd最短路）

Six Degrees of Cowvin Bacon

             Time Limit: 1000MS             Memory Limit: 65536K            Total Submissions: 3993

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 .]

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。

题解：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，我们可以枚举，最后记录最下平均值即可。

代码如下：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;const int N=350;int map[N][N];int n,m;int s[N];void init()    //初始化{    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++)            if(i!=j)                map[i][j]=1e9;   //初始化为较大的值            else                map[i][j]=0;}void Flyod()   //弗洛伊德函数{    int i,j,k;    int  minsum,sum;    long long ans=0;    //floyd-warshall 算法核心语句  松弛    for(k=1; k<=n; k++)        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                map[i][j]=min(map[i][j],map[k][j]+map[i][k]);    minsum=100000000;   //定义一个较大的最小值，然后不断更新他的最小值    for(i=1; i<=n; i++)   //枚举每一头牛    {        sum=0;        for(j=1; j<=n; j++)            sum+=map[i][j];  //记录到其他牛的度的最小值和        if(minsum>sum)   //不断更新最小值            minsum=sum;    }    cout<<minsum*100/(n-1)<<endl;   //输出其平均值的100倍}int main(){    while(cin>>n>>m)    {        int mi;        init();        for(int i=1; i<=m; i++)        {            cin>>mi;            for(int j=1; j<=mi; j++)                cin>>s[j];            for(int j=1; j<=mi; j++)  //将度入到map数组中                for(int k=j+1; k<=mi; k++)                {                    map[s[j]][s[k]]=1;                    map[s[k]][s[j]]=1;                }        }        Flyod();    }    return 0;}