<poj

　　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 = 1; k <= V; ++k) {
        for (int i = 1; i <= V; ++i) {
            for (int j = 1; 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, 0, sizeof(p));
    for (int i = 1;i <= V; ++i)
            for (int j = 1; j <= V; ++j)
                cost[i][j] = INF;
    for (int i = 1; i <= V; ++i)
        cost[i][i] = 0;
    //-----------------------------------------整理输入数据
    
    while (n--) {
        cin >> y;
        for (int i = 1; i <= y; ++i) {
            cin >> p[i];
        }
        for (int i = 1;i <= y; ++i) {
            for (int j = i + 1; j <= y; ++j) {
                a = p[i];
                b = p[j];
                cost[b][a] = cost[a][b] = 1;///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 = 1; i <= V; ++i) {
        sum = 0;
        for (j = 1; j <= V; ++j) {
                sum += cost[i][j];
        }
        if (sum < minsum)//求最小值
            minsum = sum;
    }

    cout << (minsum * 100) / (V - 1) << endl;    //输出 最小 平均值

    return 0;
}

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