题目链接：ZOJ 1718 POJ 2031 Building a Space Station 修建空间站
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
如果你是空间站工程队的一员，被分配到建设空间站的任务中，希望编写一个程序完成这个任务。
空间站由许多单元组成，这些单元被称为单间。所有的单间都是球形的，且大小不比一致。在空间站成功的进入轨道后每个单间被固定在预定的位置上。很奇怪的是，两个单间可以接触，甚至可以重叠，在极端的情况下，一个单间甚至可以完全包含另一个单间。
所有的单间都必须连接，因为宇航员可以从一个单间走到另一个单间。在以下情形，宇航员可以从单间A走到单间B。
（1）A和B接触，或者重叠。
（2）A和B用一个走廊连接。
（3）存在单间C，是的从A走到C，也可以从B走到C。
如果要求安排设计空间站的结构，也就是安排哪些单间需要用走廊连接。在设计走廊的结构时有一定的自由性。修建走廊的费用正比于它的长度。因此，需要选择一个方案，使得走廊的总长度最短。走廊修建在两个单间的表面，且假设任意两条走廊都不交叉。
分析：
首先求出所有的单间两两是否接触如果接触，说明他们之间的边长为0，不接触那么就是圆心的距离减去两个单间的半径。然后构图求最小生成树，采用Kruskal算法。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxm 5050
#define maxn 110
int m, parent[maxn];
double ans;
struct ball
{
double x, y, z, r;
}BL[maxn];
double dis(ball a, ball b)
{
return 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;
}
struct edge
{
int u, v;
double w;
}EG[maxm];
bool cmp(edge a, edge b)
{
return a.w < b.w;
}
int Find(int x)
{
if(parent[x] == -1) return x;
return Find(parent[x]);
}
void Kruskal()
{
memset(parent, -1, sizeof(parent));
ans = 0;
sort(EG, EG+m, cmp);
for(int i = 0; i < m; i++)
{
int t1 = Find(EG[i].u), t2 = Find(EG[i].v);
if(t1 != t2)
{
ans += EG[i].w;
parent[t1] = t2;
}
}
}
int main()
{
int n;
while(scanf("%d", &n), n)
{
for(int i = 0; i < n; ++i)
scanf("%lf%lf%lf%lf", &BL[i].x, &BL[i].y, &BL[i].z, &BL[i].r);
m = 0;
for(int i = 0; i < n; ++i)
for(int j = i+1; j < n; ++j)
{
double d = dis(BL[i], BL[j]);
EG[m].u = i;
EG[m].v = j;
if(d > 0)
EG[m].w = d;
else
EG[m].w = 0;
++m;
}
Kruskal();
printf("%.3lf\n", ans);
}
return 0;
}
ZOJ 1718 POJ 2031 Building a Space Station 修建空间站 最小生成树 Kruskal算法
原文地址：http://blog.csdn.net/u011439796/article/details/39404409
