poj 2139 奶牛拍电影问题 floyd算法

题意：奶牛拍一系列电影，n头牛拍m部电影，同一部电影种的搭档们距离为1，求最小距离？

思路:Floyd 图 最短路径 

1. 存图： 初始化图

   for (int i = 0; i <= n; i++)
   {
   for (int j = 0; j <= n; j++)
   f[i][j] = INF;
   f[i][i] = 0;
   }     

2. 存图：

   for (int i = 0; i < k; i++)
   {
   scanf("%d", &a[i]);
   for (int j = 0; j < i; j++)
   f[a[i]][a[j]] = f[a[j]][a[i]] = 1;
   }

3. floyd算法：

   void floyd()
   {
   for (int i = 1; i <= n; i++)
   for (int j = 1; j <= n; j++)
   for (int k = 1; k <= n; k++)
   if (f[j][k] > (f[j][i] + f[i][k]))
   f[j][k] = f[j][i] + f[i][k];
   }   解释一下a[i][j]表示i到j的距离  a[i][k]+a[k][j]表示 i到k到j的距离

4. 找到其中的一条最短路径即可

解决问题的代码：

#include <iostream>
#include <cstdio>
using namespace std;
#define INF 1<<29
#define maxn 305
int f[maxn][maxn];
int a[maxn];
int n, m;
void floyd()
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= n; k++)
                if (f[j][k] > (f[j][i] + f[i][k]))
                    f[j][k] = f[j][i] + f[i][k];
}
int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= n; j++)
                f[i][j] = INF;
            f[i][i] = 0;
        }
        while (m--)
        {
            int k;
            scanf("%d", &k);
            for (int i = 0; i < k; i++)
            {
                scanf("%d", &a[i]);
                for (int j = 0; j < i; j++)
                    f[a[i]][a[j]] = f[a[j]][a[i]] = 1;
            }
        }
        floyd();
        int ans = INF;
        for (int i = 1; i <= n; i++)
        {
            int res = 0;
            for (int j = 1; j <= n; j++)
                res += f[i][j];
            if (res < ans) ans = res;
        }
        printf("%d\n", ans * 100 / (n - 1));
    }
    return 0;
}