UVa 821 网页跳跃（Floyd）

https://vjudge.net/problem/UVA-821

题意：
给出一个有向图，任意两点都可相互到达，求任意两点的最短距离的平均值。

思路：
求两点的最短距离，用Floyd算法很方便，最后加起来算个平均值即可。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;

 const int INF = ;

 int x, y;
 int d[][];

 void Floyd()
 {
     for (int k = ; k <= ;k++)
     for (int i = ; i <= ;i++)
     for (int j = ; j <= ; j++)
         d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
 }

 int main()
 {
     //freopen("D:\\txt.txt", "r", stdin);
     int kase = ;
     while (cin >> x >> y && (x!= || y!=))
     {
         for (int i = ; i <= ;i++)
         for (int j = ; j <= ;j++)
         if (i == j)  d[i][j] = ;
         else d[i][j] = INF;

         do
         {
             d[x][y] = ;
         } while (cin >> x >> y && (x!= || y!=));
         Floyd();
         double ans = ;
         int count = ;
         for (int i = ; i <= ;i++)
         for (int j = ; j <= ; j++)
         {
             if (i!=j && d[i][j] < INF)
             {
                 count++;
                 ans += d[i][j];
             }
         }
         printf("Case %d: average length between pages = %.3f clicks\n", ++kase, ans / count);
     }
 }