题目链接:http://poj.org/problem?id=2031
Time Limit: 1000MS Memory Limit: 30000K

Description

You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task. 
The space station is made up with a number of units, called cells. All cells are sphere-shaped, but their sizes are not necessarily uniform. Each cell is fixed at its predetermined position shortly after the station is successfully put into its orbit. It is quite strange that two cells may be touching each other, or even may be overlapping. In an extreme case, a cell may be totally enclosing another one. I do not know how such arrangements are possible. 

All the cells must be connected, since crew members should be able to walk from any cell to any other cell. They can walk from a cell A to another cell B, if, (1) A and B are touching each other or overlapping, (2) A and B are connected by a `corridor', or (3) there is a cell C such that walking from A to C, and also from B to C are both possible. Note that the condition (3) should be interpreted transitively. 

You are expected to design a configuration, namely, which pairs of cells are to be connected with corridors. There is some freedom in the corridor configuration. For example, if there are three cells A, B and C, not touching nor overlapping each other, at least three plans are possible in order to connect all three cells. The first is to build corridors A-B and A-C, the second B-C and B-A, the third C-A and C-B. The cost of building a corridor is proportional to its length. Therefore, you should choose a plan with the shortest total length of the corridors. 

You can ignore the width of a corridor. A corridor is built between points on two cells' surfaces. It can be made arbitrarily long, but of course the shortest one is chosen. Even if two corridors A-B and C-D intersect in space, they are not considered to form a connection path between (for example) A and C. In other words, you may consider that two corridors never intersect. 

Input

The input consists of multiple data sets. Each data set is given in the following format. 

n 
x1 y1 z1 r1 
x2 y2 z2 r2 
... 
xn yn zn rn 

The first line of a data set contains an integer n, which is the number of cells. n is positive, and does not exceed 100. 

The following n lines are descriptions of cells. Four values in a line are x-, y- and z-coordinates of the center, and radius (called r in the rest of the problem) of the sphere, in this order. Each value is given by a decimal fraction, with 3 digits after the decimal point. Values are separated by a space character. 

Each of x, y, z and r is positive and is less than 100.0. 

The end of the input is indicated by a line containing a zero. 

Output

For each data set, the shortest total length of the corridors should be printed, each in a separate line. The printed values should have 3 digits after the decimal point. They may not have an error greater than 0.001. 

Note that if no corridors are necessary, that is, if all the cells are connected without corridors, the shortest total length of the corridors is 0.000. 

Sample Input

3
10.000 10.000 50.000 10.000
40.000 10.000 50.000 10.000
40.000 40.000 50.000 10.000
2
30.000 30.000 30.000 20.000
40.000 40.000 40.000 20.000
5
5.729 15.143 3.996 25.837
6.013 14.372 4.818 10.671
80.115 63.292 84.477 15.120
64.095 80.924 70.029 14.881
39.472 85.116 71.369 5.553
0

Sample Output

20.000
0.000
73.834

Problem solving report:

Description: 假设你是太空工作站的一员，现在安排你去完成一个任务：在N个小空间站（为球形）之间修路使他们能够互通，要求修路的长度要最小。如果两个空间站紧挨在一起，则不需要修路。多组数据输入。第一行：一个数字N，表示有N个子空间站；第2---N+1行：每行4个数据，前3个为空间站的球心，第4个为半径。
Problem solving: 很明显是最小生成树问题，但关键是如何找到边关系。我们可以找一个结构数组存放空间站信息(球心坐标和半径),每输入一个新空间站信息时，把它和之前存过的空间站相连，求出要修路的长度。由于每两个空间站在数组里是不同位置，所以在数组的序号可以看成两个点，将这两点要修路的长度看成一条边的信息存放到一个数组中，然后跑一遍最小生成树就行了。

Accepted Code:

/* 
 * @Author: lzyws739307453 
 * @Language: C++ 
 */
#include <queue>
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 105;
const int MAXM = MAXN * MAXN;
const int inf = 0x3f3f3f3f;
bool vis[MAXN];
double mp[MAXN][MAXN], dis[MAXN];
struct Circle {
    double x, y, z, r;
}p[MAXN];
struct edge {
    int v;
    double w;
    edge() {}
    edge(int v, double w) : v(v), w(w) {}
    bool operator < (const edge &s) const {
        return s.w < w;
    }
};
double prim(int s, int n) {
    priority_queue <edge> Q;
    for (int i = 0; i < n; i++) {
        vis[i] = 0;
        dis[i] = inf;
    }
    dis[s] = 0;
    Q.push(edge(s, dis[s]));
    double ans = 0;
    while (!Q.empty()) {
        edge u = Q.top();
        Q.pop();
        if (vis[u.v])
            continue;
        vis[u.v] = 1;
        ans += u.w;
        for (int j = 0; j < n; j++) {
            if (!vis[j] && dis[j] > mp[u.v][j]) {
                dis[j] = mp[u.v][j];
                Q.push(edge(j, dis[j]));
            }
        }
    }
    return ans;
}
double dist(Circle a, Circle b) {
    double t = sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y) + (a.z - b.z) * (a.z - b.z)) - a.r - b.r;
    if (t < 0)
        return 0;
    return t;
}
int main() {
    int n;
    while (scanf("%d", &n), n) {
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z, &p[i].r);
            for (int j = 0; j < i; j++)
                mp[i][j] = mp[j][i] = dist(p[i], p[j]);
        }
        printf("%.3f\n", prim(0, n));
    }
    return 0;
}