Building a Space Station

链接

POJ_2031 Building a Space Station

Describe

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 AAA to another cell BBB, if, (1) AAA and BBB are touching each other or overlapping, (2) AAA and B are connected by a `corridor’, or (3) there is a cell CCC such that walking from AAA to CCC, and also from BBB to CCC 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 AAA, BBB and CCC, 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−BA-BA−B and A−CA-CA−C, the second B−CB-CB−C and B−AB-AB−A, the third C−AC-AC−A and C−BC-BC−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−BA-BA−B and C−DC-DC−D intersect in space, they are not considered to form a connection path between (for example) AAA and CCC. 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.

nx1y1z1r1x2y2z2r2...xnynznrnn\\ x_1\ y_1\ z_1\ r_1\\ x_2\ y_2\ z_2\ r_2\\ ...\\ x_n\ y_n\ z_n\ r_nnx1 y1 z1 r1x2 y2 z2 r2...xn yn zn rn

The first line of a data set contains an integer nnn, which is the number of cells. n is positive, and does not exceed 100100100.

The following n lines are descriptions of cells. Four values in a line are xxx-, yyy- and zzz-coordinates of the center, and radius (called rrr 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,zx, y, zx,y,z and rrr is positive and is less than 100.0100.0100.0.

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 333 digits after the decimal point. They may not have an error greater than 0.0010.0010.001.

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 0.0000.0000.000.

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

解析

几个要点：
1、三维空间内两点坐标公式与二维空间内相比只是多了一个 zzz 坐标的平方；
2、两空间站之间“走廊”的长度是两空间站中心直线距离减去各自的半径 rrr；
3、由要点2易得，判断两空间站是否重合或相交只需要看求出的走廊长度是否为非正整数即可；
4、Kruskal算法要配合并查集算法一起使用；

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
struct str
{int a,b;double val;
}edg[20001];
struct xyz
{double x,y,z,r;
}a[201];
int cnt;
double ans;
int n,pre[201],fx,fy;
void add(int frm,int to,double val)
{cnt++;edg[cnt].a=frm;edg[cnt].b=to;edg[cnt].val=val;
}
bool cmp(str a,str b)
{return a.val<b.val;
}
double dis(xyz a,xyz b)
{double xx,yy,zz,tmp;xx=(a.x-b.x)*(a.x-b.x);yy=(a.y-b.y)*(a.y-b.y);zz=(a.z-b.z)*(a.z-b.z);tmp=sqrt(xx+yy+zz);if(tmp-(a.r+b.r)>0)return tmp-(a.r+b.r);elsereturn 0;
}
int find_father(int x)
{if(pre[x]==x)return x;elsereturn pre[x]=find_father(pre[x]);
}
int main()
{while(1){scanf("%d",&n);if(!n)break;ans=0,cnt=0;for(int i=1;i<=n;i++)pre[i]=i;for(int i=1;i<=n;i++)cin>>a[i].x>>a[i].y>>a[i].z>>a[i].r;for(int i=1;i<=n-1;i++)for(int j=i+1;j<=n;j++)if(!dis(a[i],a[j])){fx=find_father(i);fy=find_father(j);if(fx!=fy)pre[fy]=fx;}elseadd(i,j,dis(a[i],a[j]));sort(edg+1,edg+cnt+1,cmp);for(int i=1;i<=cnt;i++){fx=find_father(edg[i].a);fy=find_father(edg[i].b);if(fx!=fy){pre[fy]=fx;ans+=edg[i].val;}}printf("%.3f\n",ans);}return 0;
}

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