题意：三维空间中n个球要相连。。。通路的代价是距离。。。如果球相交（切）或者包含那么不用建通路就能联系。。。问联系所有球的最小代价。。。
思路：裸的最小生成树。。。。先预处理球和球表面的距离。。。距离是负数的处理成0.。。然后mst搞之。。。不算CE的话是1A….
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年07月14日 星期四 15时44分53秒
4File Name :code/poj/2031.cpp
5 ************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31const int N =105;
32int n;
33int m;
34double a[N][N];
35int f[N];
36int dblcmp(double d)
37{
38 return d<-eps?-1:d>eps;
39}
40void init()
41{
42 for ( int i = 1 ; i <= n ; i++) f[i] = i;
43}
44struct circle
45{
46 double x,y,z,r;
47 double dis(circle b)
48 {
49 double res = 0 ;
50 res = (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y)+(z-b.z)*(z-b.z);
51 res = sqrt(res);
52 res = res - r - b.r;
53 if (dblcmp(res)<0) res = 0 ;
54 return res;
55 }
56 void input()
57 {
58 scanf("%lf%lf%lf%lf",&x,&y,&z,&r);
59 }
60}cir[N];
61struct Edge
62{
63 int u,v;
64 double w;
65 bool operator < (Edge b)const
66 {
67 return w<b.w;
68 }
69}edge[N*N/2];
70int root ( int x)
71{
72 if (x!=f[x]) f[x] = root (f[x]);
73 return f[x];
74}
75void merge( int x,int y)
76{
77 int rx = root (x);
78 int ry = root(y);
79 if (rx==ry) return;
80 f[rx] = ry;
81}
82int main()
83{
84#ifndef ONLINE_JUDGE
85 freopen("code/in.txt","r",stdin);
86#endif
87 while (~scanf("%d",&n))
88 {
89 if (n==0) break;
90 init();
91 for ( int i = 1 ;i <= n ; i++) cir[i].input();
92 for ( int i = 1 ; i <= n ; i++)
93 for ( int j = 1 ; j <= n ; j++)
94 a[i][j]=a[j][i]=cir[i].dis(cir[j]);
95 m = 0 ;
96 for ( int i = 1 ; i <= n ; i++)
97 for ( int j = 1 ; j <= n ; j++)
98 if (i<j)
99 {
100 m++;
101 edge[m].u = i ;
102 edge[m].v = j;
103 edge[m].w = a[i][j];
104 }
105 sort(edge+1,edge+m+1);
106 double mst = 0 ;
107 int cnt = 0 ;
108 for ( int i = 1 ; i <= m ; i++)
109 {
110 int u = edge[i].u;
111 int v = edge[i].v;
112 double w = edge[i].w;
113 if (root(u)==root(v)) continue;
114 merge(u,v);
115 mst +=w;
116 cnt++;
117 if (cnt>=n-1) break;
118 }
119 if (dblcmp(mst)==0)
120 {
121 puts("0.000");
122 }else
123 {
124 printf("%.3f\n",mst);
125 }
126 }
127#ifndef ONLINE_JUDGE
128 fclose(stdin);
129#endif
130 return 0;
131}