<poj - 2139> Six Degrees of Cowvin Bacon 最短路径问题 the cow have been making movies

　　本题链接：http://poj.org/problem?id=2139

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：有一群牛在拍电影（牛们的生活真丰富），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编号。问哪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。

　解题思路：只要把任意两个牛之间的最小度求出来然后就处理就行了，可以考虑用Floyd算法（三个for）。

　参考代码：

 #include <cstring>
 #include <iostream>
 #define INF 9999999
 #define maxn 300
 using namespace std;

 int cost[maxn][maxn];
 int p[maxn];
 int V;

 void fl () {
     for (int k = ; k <= V; ++k) {
         for (int i = ; i <= V; ++i) {
             for (int j = ; j <= V; ++j) {
                 cost[i][j] = min (cost[i][j], cost[i][k] + cost[k][j]);
             }
         }
     }
 }

 int main () {
     int x, y;
     int a, b;
     int n;

     cin >> V >> n;

     //-----------------------------------------初始化
     memset (p, , sizeof(p));
     for (int i = ;i <= V; ++i)
             for (int j = ; j <= V; ++j)
                 cost[i][j] = INF;
     for (int i = ; i <= V; ++i)
         cost[i][i] = ;
     //-----------------------------------------整理输入数据

     while (n--) {
         cin >> y;
         for (int i = ; i <= y; ++i) {
             cin >> p[i];
         }
         for (int i = ;i <= y; ++i) {
             for (int j = i + ; j <= y; ++j) {
                 a = p[i];
                 b = p[j];
                 cost[b][a] = cost[a][b] = ;///a和b之间的距离
             }
         }
     }

     fl ();//--------------调用函数

     //-------------------------------------------------------求最小值输出
     int sum, minsum = INF;
     int i, j;

     for (i = ; i <= V; ++i) {
         sum = ;
         for (j = ; j <= V; ++j) {
                 sum += cost[i][j];
         }
         if (sum < minsum)//求最小值
             minsum = sum;
     }

     cout << (minsum * ) / (V - ) << endl;    //输出 最小 平均值

     return ;
 }

　　欢迎码友评论，一起成长。