POJ Building a Space Station 最小生成树
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 15664 | Accepted: 6865 |
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
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
这个题主要还是考察建图,将空间站的形状看成球即可。两空间站的cell 之间的距离应该为空间距离减去两空间站cell的半径,即为sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)-r1-r2。剩下的就是求一颗最新小生成树,因为在做最小生成树专题,对于题目的判断还是不能锻炼,这里告诉你了是最小生成树,顺着这个思路向下去想,很容易就找到了解决的思路。稠密图,直接Prim盘他
#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//
#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define debug(n) printf("%d_________________________________\n",n);
#define speed ios_base::sync_with_stdio(0)
#define file freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) (a>b?a:b)
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
#define pb(n) push_back(n)
#define dis(a,b,c,d) ((double)sqrt((a-c)*(a-c)+(b-d)*(b-d)))
//--------------------------------constant----------------------------------//
#define INF 0x3f3f3f3f
#define esp 1e-9
#define PI acos(-1)
using namespace std;
struct node
{
double x,y,z,r;
int i;
}f[101];
double low[101];
double G[101][101];
void calc(const node &a,const node &b){
double dis=sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2)+pow(a.z-b.z,2));
if(dis<a.r+b.r)
G[a.i][b.i]=G[b.i][a.i]=0;
else
G[a.i][b.i]=G[b.i][a.i]=dis-(a.r+b.r);
}
double prim(int n)
{
double ans=0,min;
bool vis[101]={0};
int pos=0,i,j;
for(int i=0;i<n;++i)
low[i]=G[0][i];
vis[pos]=true;
for(i=1;i<n;++i ){
for(j=0,min=low[j],pos=j;j<n;++j)if(!vis[j]){
if(low[j]<min){
min=low[j];
pos=j;
}
}
ans+=min;
vis[pos]=true;
for(j=0;j<n;++j){
if(!vis[j]&&low[j]>G[pos][j]){
low[j]=G[pos][j];
}
}
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n),n){
for(int i=0;i<n;++i){
scanf("%lf %lf %lf %lf",
&f[i].x,&f[i].y,&f[i].z,&f[i].r);
f[i].i=i;
}
for(int i=0;i<n;++i)
for(int j=0;j<n;++j)if(i==j)
G[i][j]=100000000.00;//给double赋值一定要.000不然WA
for(int j=0;j<n;++j)
for(int i=j+1;i<n;++i){
calc(f[j],f[i]);
}
printf("%.3lf\n",prim(n));
}
}
POJ Building a Space Station 最小生成树的更多相关文章
- POJ 2031:Building a Space Station 最小生成树
Building a Space Station Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6083 Accepte ...
- 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 ...
- POJ - 2031C - Building a Space Station最小生成树
You are a member of the space station engineering team, and are assigned a task in the construction ...
- POJ 2031 Building a Space Station 最小生成树模板
题目大意:在三维坐标中给出n个细胞的x,y,z坐标和半径r.如果两个点相交或相切则不用修路,否则修一条路连接两个细胞的表面,求最小生成树. 题目思路:最小生成树树模板过了,没啥说的 #include& ...
- poj Building a Space Station
http://poj.org/problem?id=2031 #include<cstdio> #include<cstring> #include<cmath> ...
- 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 ...
- 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 ...
- 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 ...
- Building a Space Station POJ - 2031
Building a Space Station POJ - 2031 You are a member of the space station engineering team, and are ...
随机推荐
- 总结关于Mac上使用MySQL一些常见的问题
Num 1. MySQL5.7导出数据时提示--secure-file-priv解决办法: 问题分析 在官方的文档中,对secure_file_priv进行了说明,它用于限制数据的导出. secur ...
- redis 练习 a的数据库数据迁移到b数据库
思路 1.从a redis中获取所有的key 2.判断key的类型 3.根据key的类型,判断使用的是set/hset类型 4.set到b redis中(写入到b redis中)
- Git中rebase失败了如何进行恢复
rebase失败后的恢复 记一次翻车现场 记一次翻车的现场,很早之前提的PR后面由于需求的变便去忙别的事情了,等到要做这个需求的我时候,发现已经 落后版本了,并且有很多文件的冲突,然后就用rebase ...
- elasticsearch7.6.2实战(2)-es可视化及分析平台-kibana
1. 场景描述 elasticsearch部署完成后,es官方提供了可视化.分析及管理平台-kibana,部署下,有需要朋友参考下,不谢! 2. 解决方案 2.1 下载 (1)地址:https://w ...
- svg如何设置中心点进行缩放
中心点设置:x = x+width/2 y=y+height/2缩放开始前后需要变换对应的位置,直接举例:<rect x="110" y="100" ...
- 面试问了解Linux内存管理吗?10张图给你安排的明明白白!
文章每周持续更新,各位的「三连」是对我最大的肯定.可以微信搜索公众号「 后端技术学堂 」第一时间阅读(一般比博客早更新一到两篇) 今天来带大家研究一下Linux内存管理.对于精通 CURD 的业务同学 ...
- 教你如何在工作中“偷懒”,python优雅的帮你解决
前言 本文的文字及图片来源于网络,仅供学习.交流使用,不具有任何商业用途,版权归原作者所有,如有问题请及时联系我们以作处理. PS:如有需要Python学习资料的小伙伴可以加点击下方链接自行获取htt ...
- GCD - Extreme (II) UVA - 11426 欧拉函数与gcd
题目大意: 累加从1到n,任意两个数的gcd(i,j)(1=<i<n&&i<j<=n). 题解:假设a<b,如果gcd(a,b)=c.则gcd(a/c,b ...
- 原创hadoop2.6集群环境搭建
三台机器: Hmaster 172.168.2.3.Hslave1 172.168.2.4.Hslave2 172.168.2.6 JDK:1.8.49 OS:red hat 5.4 64 (由于后期 ...
- ES6新增的 Set 和 WeakSet 是什么玩意?在此揭晓
现在的章节内容会更加的紧密,如果大家看不懂可以先去看以前的文章,当然看了的忘了,也可以去看一下,这样学习后面的内容才会更加容易. 什么是Set结构 Set是ES6给开发者带来的一种新的数据结构,你可以 ...