【POJ】2031 Building a Space Station

题目链接：http://poj.org/problem?id=2031

题意：修建太空站每个舱之间的走廊。每个舱都是球体。给出n个舱的三维空间坐标以及球体半径。如果球体之间接触或者相接，就不用修走廊。让你求最短走廊的长度。

题解：有点点坑这个题。。改了好久。。这里的存储其实是 $len(a,b) - r_a - r_b$ 求一下这个存储再以Prim求解即可。注意精度。

代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 const int maxn = ;
 const int inf = 0x3f3f3f3f;
 const double eps = 1e-;
 double mp[maxn][maxn];
 double dis[maxn];
 bool vis[maxn];
 int n;

 struct node{
     double x;
     double y;
     double z;
     double r;
 };

 node cir[maxn];

 double dist(node a,node b){
     double diss =  (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) + (a.z -b.z) * (a.z - b.z);
     return sqrt(diss) - a.r - b.r;
 }

 double prim(){
     double ans =  ;
     int k;
     double minn ;
     for(int i =  ; i <= n ; i++){
         dis[i] = inf ;
         vis[i] = false ;
     }
     dis[] =  ;
     for(int i =  ; i <= n ; i++){
         minn = inf ;
         k =  ;
         for(int j =  ; j <= n ; j++){
             if(minn > dis[j] && !vis[j]){
                 minn = dis[j] ;
                 k = j ;
             }
         }
         if(minn >= inf)
             return - ;
         ans += minn ;
         vis[k] = true ;
         for(int j =  ; j <= n ; j++){
             if(!vis[j] && dis[j] > mp[k][j])
                 dis[j] = mp[k][j] ;
         }
     }
     return ans ;
 }

 int main(){
     while(cin>>n && n){
         for(int i = ; i <= n ;i++){
             cin>>cir[i].x>>cir[i].y>>cir[i].z>>cir[i].r;
         }
         memset(mp,,sizeof(mp));
         for(int i = ; i < n; i++){
             for(int j = i+; j <= n ;j++){
                 if(dist(cir[i],cir[j]) < eps){
                     mp[i][j] = mp[j][i] = ;
                 }
                 else{
                     mp[i][j] = mp[j][i] = dist(cir[i],cir[j]);
                 }
             }
         }
         /*
         for(int i = 1; i <= n; i++){
             for(int j = 1 ; j <= n ;j++){
                 cout<<mp[i][j]<<" ";
             }
             cout<<endl;
         }*/
         double ans = prim();
         if(ans != -){
             printf("%.3lf\n",ans);
         }

     }

     return ;
 }