3维空间中的最小生成树。。。。
好久没碰关于图的东西了。。。。。

             Building a Space Station
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 3804   Accepted: 1940

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.


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

Source

Japan 2003 Domestic

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath> using namespace std; const double eps=1e-;
const double INF=0x3f3f3f3f; struct node
{
double x,y,z,r;
node() {}
node(double a,double b,double c,double d):x(a),y(b),z(c),r(d) {}
}space[]; double Dist(node a,node b)
{
double ans=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));
ans=ans-a.r-b.r;
if(ans<eps) ans=;
return ans;
} double d[][]; int n;
bool vis[];
double dis[]; double PRIM()
{
memset(vis,false,sizeof(vis));
for(int i=;i<n;i++) dis[i]=INF;
dis[]=;double ans=;
for(int i=;i<n;i++)
{
int mark=-;
for(int j=;j<n;j++)
{
if(!vis[j])
{
if(mark==-) mark=j;
else if(dis[j]<dis[mark]) mark=j;
}
}
if(mark==-) break;
vis[mark]=true;
ans+=dis[mark];
for(int j=;j<n;j++)
{
if(j==mark) continue;
if(!vis[j])
{
dis[j]=min(dis[j],d[mark][j]);
}
}
}
return ans;
} int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
for(int i=;i<n;i++)
{
double a,b,c,d;
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
space[i]=node(a,b,c,d);
}
memset(dis,,sizeof(dis));
for(int i=;i<n;i++)
{
for(int j=;j<n;j++)
{
if(i==j) continue;
d[i][j]=Dist(space[i],space[j]);
}
}
printf("%.3lf\n",PRIM());
}
return ;
}

POJ 2031 Building a Space Station的更多相关文章

  1. POJ 2031 Building a Space Station【经典最小生成树】

    链接: http://poj.org/problem?id=2031 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#probl ...

  2. poj 2031 Building a Space Station【最小生成树prime】【模板题】

    Building a Space Station Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5699   Accepte ...

  3. POJ 2031 Building a Space Station (最小生成树)

    Building a Space Station Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 5173   Accepte ...

  4. POJ 2031 Building a Space Station (最小生成树)

    Building a Space Station 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/C Description Yo ...

  5. POJ - 2031 Building a Space Station 三维球点生成树Kruskal

    Building a Space Station You are a member of the space station engineering team, and are assigned a ...

  6. POJ 2031 Building a Space Station (计算几何+最小生成树)

    题目: Description You are a member of the space station engineering team, and are assigned a task in t ...

  7. POJ 2031 Building a Space Station【最小生成树+简单计算几何】

    You are a member of the space station engineering team, and are assigned a task in the construction ...

  8. POJ 2031 Building a Space Station (prim裸题)

    Description You are a member of the space station engineering team, and are assigned a task in the c ...

  9. poj 2031 Building a Space Station(prime )

    这个题要交c++, 因为prime的返回值错了,改了一会 题目:http://poj.org/problem?id=2031 题意:就是给出三维坐标系上的一些球的球心坐标和其半径,搭建通路,使得他们能 ...

随机推荐

  1. visual studio 2010 C#编程时 没有.NET framework 2.0目标框架的解决办法

    解决办法是安装Framework .NET 3.5 Sp1 因为visual studio 2010是依赖.NET Framework 3.5 Sp1来识别其它版本的.NEt framework的. ...

  2. Session对象

    Session对象用于存储在多个页面调用之间特定用户的信息.Session对象只针对单一网站使用者,不同的客户端无法互相访问.Session对象中止于联机机器离线时,也就是当网站使用者关掉浏览器或超过 ...

  3. Oracle 数据类型

    类型 含义 CHAR(length) 存储固定长度的字符串.参数length指定了长度,如果存储的字符串长度小于length,用空格填充.默认长度是1,最长不超过2000字节. VARCHAR2(le ...

  4. POJ 2533 Longest Ordered Subsequence(最长上升子序列(NlogN)

    传送门 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subseque ...

  5. 第二次作业———“A+B Format”思路与总结

    GitHub链接: https://github.com/zzy19961112/object-oriented "A+B Format" 题目 解题思路: 一开始粗略看这道题,熟 ...

  6. Compiler Warning (level 3) C4800

    #pragma warning( disable : 4800 ) // forcing bool 'true' or 'false' ,忽略4800 警告#pragma warning( disab ...

  7. Possion 分布

    泊松分布的概率函数为: \[P(X=k)=\frac{\lambda^k}{k!}e^{-\lambda},k=0,1,2,\cdots\] 如果 $X_i  \sim P(\lambda_i)$,并 ...

  8. zabbix监控系列(4)之zabbix报警邮件无法发送

    情况介绍 首先确保邮箱规则没有把报警邮件作为垃圾邮件拉黑了. 服务器断电重启后,发现zabbix报警邮件无法发送,断电之前是好好的,但是重启后不行了,于是查看maillog日志,发现这个错误: Hos ...

  9. UVA 11419SAM I AM(输出 最小覆盖点 )

    参考博客:如何找取 最小覆盖点集合 题意:R*C大小的网格,网格上面放了一些目标.可以再网格外发射子弹,子弹会沿着垂直或者水平方向飞行,并且打掉飞行路径上的所有目标,计算最小多少子弹,各从哪些位置发射 ...

  10. webservice理解

    什么是webservice? 1.基于web的一种服务,webservice分为服务器端server和客户端client. server端会会提供一些资源供客户端的应用来访问(获取所需要的数据) 2. ...