UVA821 floyd最短路+暴力

题意：给n条边，求每两个点之间的平均距离；

思路：数据是100条边，用floyd得到每两点之间的最短距离，然后遍历相加除以边的数目；

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #define N 110
 #define INF 0x7ffffff
 #define repu(i,a,b) for(int i=a;i<b;i++)
 using namespace std;
 int p[N][N],vis[N][N];
 int floyd(int n,int m)
 {
     repu(k,n,m)
     {
         repu(i,n,m)
         repu(j,n,m)
         if(p[i][k] < INF && p[k][j] < INF)
         {
             p[i][j] = min(p[i][j],p[i][k] + p[k][j]);
         }
     }
 }
 int main()
 {
     int u,v,kase=;
     while()
     {
         repu(i,,N)
         {
             repu(j,,N)
             p[i][j] = INF,vis[i][j] = ;
             p[i][i] = ;
         }
         memset(vis,,sizeof(vis));
         scanf("%d%d",&u,&v);
         if(!u&&!v)
             break;
         p[u][v] = ;
         vis[u][v] = ;
         int minn = INF,maxn = ;
         while()
         {
             scanf("%d%d",&u,&v);
             if(!u&&!v)
                 break;
             p[u][v] = ;
             if(!vis[u][v])
                 vis[u][v] = ;
             maxn = max(max(v,u),maxn);
             minn = min(min(v,u),minn);
         }
         int sum = ,t = ;
         floyd(minn,maxn+);
         repu(i,minn,maxn+)
         {
             repu(j,minn,maxn+)
             if(p[i][j]!=INF && i!=j )
                 t++,sum += p[i][j];
         }
         double q = (double)sum / (double)t;
        // cout<<t<<endl<<sum<<endl;
         printf("Case %d: average length between pages = %.3lf clicks\n",kase++,q);
     }
     return ;
 }