POJ-2031 Building a Space Station （球的最小生成树）

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

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

10.000 10.000 50.000 10.000

40.000 10.000 50.000 10.000

40.000 40.000 50.000 10.000

30.000 30.000 30.000 20.000

40.000 40.000 40.000 20.000

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

Sample Output

题意：

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个为空间站的球心的坐标x,y,z，第4个为半径r

思路：

很明显是最小生成树问题，关键是如何找到边关系

如果球之间相交，那么距离为0，否则求下

注意一些数据是double，别忘了

代码如下：

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <string>

 #include <math.h>

 #include <algorithm>

 #include <vector>

 #include <stack>

 #include <queue>

 #include <set>

 #include <map>

 #include <sstream>

 const int INF=0x3f3f3f3f;

 typedef long long LL;

 const int mod=1e9+;

 //const double PI=acos(-1);

 #define Bug cout<<"---------------------"<<endl

 const int maxn=1e4+;

 using namespace std;

 struct edge_node

 {

     int to;

     double val;

     int next;

 }Edge[maxn*maxn/];

 int Head[maxn];

 int tot;

 struct point_node

 {

     double x,y,z;

     double r;

 }PT[maxn];

 void Add_Edge(int u,int v,double w)

 {

     Edge[tot].to=v;

     Edge[tot].val=w;

     Edge[tot].next=Head[u];

     Head[u]=tot++;

 }

 double lowval[maxn];

 int pre[maxn];//记录每个点的双亲是谁

 double Prim(int n,int st)//n为顶点的个数,st为最小生成树的开始顶点

 {

     double sum=;

     fill(lowval,lowval+n,INF);//不能用memset(lowval,INF,sizeof(lowval))

     memset(pre,,sizeof(pre));

     lowval[st]=-;

     pre[st]=-;

     for(int i=Head[st];i!=-;i=Edge[i].next)

     {

         int v=Edge[i].to;

         double w=Edge[i].val;

         lowval[v]=min(lowval[v],w);

         pre[v]=st;

     }

     for(int i=;i<n-;i++)

     {

         double MIN=INF;

         int k;

         for(int i=;i<n;i++)//根据编号从0或是1开始，改i从0--n-1和1--n

         {

             if(lowval[i]!=-&&lowval[i]<MIN)

             {

                 MIN=lowval[i];

                 k=i;

             }

         }

         sum+=MIN;

         lowval[k]=-;

         for(int j=Head[k];j!=-;j=Edge[j].next)

         {

             int v=Edge[j].to;

             double w=Edge[j].val;

             if(w<lowval[v])

             {

                 lowval[v]=w;

                 pre[v]=k;

             }

         }

     }

     return sum;

 }

 int main()

 {

     int n;

     while(~scanf("%d",&n)&&n!=)

     {

         memset(Head,-,sizeof(Head));

         tot=;

         for(int i=;i<n;i++)

         {

             scanf("%lf %lf %lf %lf",&PT[i].x,&PT[i].y,&PT[i].z,&PT[i].r);

         }

         for(int i=;i<n;i++)

         {

             for(int j=;j<n;j++)

             {

                 if(i!=j)

                 {

                     double x,y,z;

                     x=PT[i].x-PT[j].x;

                     y=PT[i].y-PT[j].y;

                     z=PT[i].z-PT[j].z;

                     double val=sqrt(x*x+y*y+z*z)-PT[i].r-PT[j].r;//球心距离减去两个半径的距离

                     if(val>)

                         Add_Edge(i,j,val);

                     else

                         Add_Edge(i,j,);

                 }

             }

         }

         printf("%.3f\n",Prim(n,));

     }

     return ;

 }