Building a Space Station(bfs)

http://poj.org/problem?id=2031

题意：给出n个球的圆心坐标x,y,z, 半径r，若任意两球不相交，则在两球间建桥。问需建桥的最短距离是多少。

思路：建图,以两球间相差的距离为权值，然后求最小生成树。

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 const int inf=<<;
 const double eps=1e-;
 const int maxn=;
 double sum,Map[maxn][maxn],dis[maxn];
 int vis[maxn];
 struct point
 {
     double x,y,z,r;
 };
 void prim(int n)
 {
     for (int i = ; i <= n; i++)
         dis[i]=inf;
     dis[]=;
     sum=;
     for (int i = ; i <= n; i++)
     {
         double Min=inf;
         int pos=;
         for (int j = ; j <= n; j++)
         {
             if (!vis[j]&&dis[j] < Min)
             {
                 Min = dis[j];
                 pos = j;
             }
         }
         if (Min==inf)
             return;
         vis[pos] = ;
         sum+=Min;
         for (int j = ; j <= n; j++)
         {
             if (!vis[j]&&dis[j] > Map[pos][j])
                 dis[j] = Map[pos][j];
         }
     }
 }
 void init()
 {
     sum = ;
     memset(vis,,sizeof(vis));
     memset(Map,,sizeof(Map));
 }
 double dist(point &a,point &b)
 {
     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
 }
 int main()
 {
     int n;
     while(~scanf("%d",&n)&&n)
     {
         point a[];
         init();
         for (int i = ; i <= n; i++)
         {
             scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);
         }
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j <= n; j++)
             {
                 double d = sqrt(dist(a[i],a[j]));
                 if (d-(a[i].r+a[j].r)>eps)
                     Map[i][j] = d-a[i].r-a[j].r;
                 else
                     Map[i][j] = ;
             }
         }
         prim(n);
         printf("%.3f\n",sum);
     }
     return ;
 }