博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
POJ—Building a Space Station
阅读量:6914 次
发布时间:2019-06-27

本文共 4061 字,大约阅读时间需要 13 分钟。

Building a Space Station
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 6657   Accepted: 3253

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

310.000 10.000 50.000 10.00040.000 10.000 50.000 10.00040.000 40.000 50.000 10.000230.000 30.000 30.000 20.00040.000 40.000 40.000 20.00055.729 15.143 3.996 25.8376.013 14.372 4.818 10.67180.115 63.292 84.477 15.12064.095 80.924 70.029 14.88139.472 85.116 71.369 5.5530

Sample Output

20.0000.000

73.834

 

#include
#include
#include
#include
#define MAXN 10000using namespace std;int per[MAXN];struct vl{ double x,y,z,r;}v[MAXN];struct node{ int v; int u; double len;}g[MAXN];int find(int x){ if(x==per[x]) return x; else return find(per[x]);} double ddis(vl v1,vl v2){ return (v1.x-v2.x)*(v1.x-v2.x)+(v1.y-v2.y)*(v1.y-v2.y)+(v1.z-v2.z)*(v1.z-v2.z);}bool cmp(node x1,node x2){ return x1.len

 

 

 

转载于:https://www.cnblogs.com/cniwoq/p/6770985.html

你可能感兴趣的文章
游戏安全资讯精选 2018年第三期:中国已经诞生了区块链手游,游戏公司向直播答题行业的借鉴宝典,微软“周二补丁日”一览...
查看>>
第一章:SpringCloud服务发现和服务注册
查看>>
区块链:定义未来金融与经济新格局
查看>>
Dom事件
查看>>
小册笔记
查看>>
mongoDB高级查询这一篇就够了
查看>>
js节流和防抖
查看>>
MySQL学习笔记之三排序和过滤
查看>>
VUE 使用笔记
查看>>
(转)Android studio 多渠道打包(超简洁版)
查看>>
你好!未来的我
查看>>
iOS 【奇巧淫技】获取webView内容高度
查看>>
阿里云CentOS MYSQL无法访问3306端口解决方案之一(不建议)
查看>>
java基础-多线程初步了解
查看>>
零基础微信开发之自动回复电影
查看>>
spring Cloud Gateway 入门简单使用
查看>>
SpringBoot源码解析-内嵌Tomcat容器的启动
查看>>
Flow_学习笔记
查看>>
阿里Java面试题剖析:关于系统拆分,为什么要进行系统拆分?
查看>>
Application 详解
查看>>