UVA Getting in Line
题目例如以下:
| Getting in Line |
Computer networking requires that the computers in the network be linked.
This problem considers a ``linear" network in which the computers are chainedtogether so that each is connected to exactly two others except for the two computers on the ends of the chain which are connected to only one other computer. A picture is shown
below. Here the computers are the black dots and their locations in the network are identified by planar coordinates (relative to a coordinate system not shown in the picture).
Distances between linked computers in the network are shown in feet.

For various reasons it is desirable to minimize the length of cable used.
Your problem is to determine how the computers should be connected into such a chain to minimize the total amount of cable needed. In the installation being constructed, the cabling will run beneath the floor, so the amount of cable used to join 2 adjacent
computers on the network will be equal to the distance between the computers plus 16 additional feet of cable to connect from the floor to the computers and provide some slack for ease of installation.
The picture below shows the optimal way of connecting the computers shownabove, and the total length of cable required for this configuration is (4+16)+ (5+16) + (5.83+16) + (11.18+16) = 90.01 feet.

Input
The input file will consist of a series of data sets. Each data set will begin with a line consisting of a single number indicating the number of computers in a network. Each network has at least 2 and at most 8 computers. A value of 0 for the number of
computers indicates the end of input.
After the initial line in a data set specifying the number of computers in a network, each additional line in the data set will give the coordinates of a computer in the network. These coordinates will be integers in the range 0 to 150. No two computers
are at identical locations and each computer will be listed once.
Output
The output for each network should include a line which tells the number of the network (as determined by its position in the input data), and one line for each length of cable to be cut to connect each adjacent pair of computers in the network. The final
line should be a sentence indicating the total amount of cable used.
In listing the lengths of cable to be cut,traverse the network from one end to the other. (It makes no difference atwhich end you start.) Use a format similar to the one shown in the sample output, with a line of asterisks separating output
for different networks and with distances in feet printed to 2 decimal places.
Sample Input
6
5 19
55 28
38 101
28 62
111 84
43 116
5
11 27
84 99
142 81
88 30
95 38
3
132 73
49 86
72 111
0
Sample Output
**********************************************************
Network #1
Cable requirement to connect (5,19) to (55,28) is 66.80 feet.
Cable requirement to connect (55,28) to (28,62) is 59.42 feet.
Cable requirement to connect (28,62) to (38,101) is 56.26 feet.
Cable requirement to connect (38,101) to (43,116) is 31.81 feet.
Cable requirement to connect (43,116) to (111,84) is 91.15 feet.
Number of feet of cable required is 305.45.
**********************************************************
Network #2
Cable requirement to connect (11,27) to (88,30) is 93.06 feet.
Cable requirement to connect (88,30) to (95,38) is 26.63 feet.
Cable requirement to connect (95,38) to (84,99) is 77.98 feet.
Cable requirement to connect (84,99) to (142,81) is 76.73 feet.
Number of feet of cable required is 274.40.
**********************************************************
Network #3
Cable requirement to connect (132,73) to (72,111) is 87.02 feet.
Cable requirement to connect (72,111) to (49,86) is 49.97 feet.
Number of feet of cable required is 136.99.
求连接几个点的一条有向路,使得路长最短。
因为数据较小,能够枚举几个点的全排列,分别算出路长,再求最小的那个。也能够用DFS+回溯,是更一般的方法,这种方法中用一个数组S记录节点的位置,避免设置麻烦的答案数组。
AC代码例如以下:
UVA Getting in Line的更多相关文章
- uva 11174 Stand in a Line
// uva 11174 Stand in a Line // // 题目大意: // // 村子有n个村民,有多少种方法,使村民排成一条线 // 使得没有人站在他父亲的前面. // // 解题思路: ...
- CDQ分治入门 + 例题 Arnooks's Defensive Line [Uva live 5871]
CDQ分治入门 简介 CDQ分治是一种特别的分治方法,它由CDQ(陈丹琦)神犇于09国家集训队作业中首次提出,因此得名.CDQ分治属于分治的一种.它一般只能处理非强制在线的问题,除此之外这个算法作为某 ...
- UVA 216 - Getting in Line
216 - Getting in Line Computer networking requires that the computers in the network be linked. This ...
- Getting in Line UVA 216
Getting in Line Computer networking requires that the computers in the network be linked. This pro ...
- UVA 11174 Stand in a Line 树上计数
UVA 11174 考虑每个人(t)的所有子女,在全排列中,t可以和他的任意子女交换位置构成新的排列,所以全排列n!/所有人的子女数连乘 即是答案 当然由于有MOD 要求逆. #include & ...
- UVa 12657 Boxes in a Line(应用双链表)
Boxes in a Line You have n boxes in a line on the table numbered 1 . . . n from left to right. Your ...
- uva 11174 Stand in a Line (排列组合)
UVa Online Judge 训练指南的题目. 题意是,给出n个人,以及一些关系,要求对这n个人构成一个排列,其中父亲必须排在儿子的前面.问一共有多少种方式. 做法是,对于每一个父节点,将它的儿子 ...
- Boxes in a Line UVA - 12657
You have n boxes in a line on the table numbered 1...n from left to right. Your task is to simulat ...
- UVA 12657 Boxes in a Line 双向链表
题目连接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=47066 利用链表换位置时间复杂度为1的优越性,同时也考虑到使用实际 ...
随机推荐
- CodeForces 484A Bits
意甲冠军: 10000询价 每次查询输入L和R(10^18) 在区间的二进制输出指示1大多数数字 1个数同样输出最小的 思路: YY一下 认为后几位全是1的时候能保证1的个数多 那么怎样构造 ...
- Unity3D-RPG项目实战(1):发动机的特殊文件夹
前言 从8月份開始.下定决心正式開始学习Unit3D啦.尽管自己写过两代端游引擎,被应用的项目也超过10个,Unreal Engine也搞过几年,只是做手游.哥确实还是个新手.Unity3D这个引擎我 ...
- IOS 多于UIImageView 当加载较大的高清闪存管理
当我们是一家人View 多于UIImageView,和UIImageView表明一个更大的HD,可能存在的存储器的警告的问题.假设第一次走进这个view,无记忆出现预警.当重新进入view,在那曾经 ...
- poj 2001 Shortest Prefixes(特里)
主题链接:http://poj.org/problem?id=2001 Description A prefix of a string is a substring starting at the ...
- android 原生应用、Web应用、混合应用优缺点分析
近期开发几个项目,牵涉到android的几种开发模式.对于原生态开发.web 应用开发以及混合模式开发,本人觉得并非哪一种就是最好的,哪一种就是最差的,这个全然是依据项目的实际需求,选择一种合适的开发 ...
- fzu 2150 Fire Game 【身手BFS】
称号:fzupid=2150"> 2150 Fire Game :给出一个m*n的图,'#'表示草坪,' . '表示空地,然后能够选择在随意的两个草坪格子点火.火每 1 s会向周围四个 ...
- Objective-c正确的写法单身
Singleton模式iOS发展可能是其中最常用的模式中使用的.但是因为oc语言特性本身,想要写一个正确的Singleton模式是比较繁琐,iOS中单例模式的设计思路. 关于单例模式很多其它的介绍请參 ...
- Net 项目构建基于Jenkins + Github + Mono 的持续集成环境
Net 项目构建基于Jenkins + Github + Mono 的持续集成环境 阅读目录 1 安装 2 配置 3 测试 在Redhat enterprise 6.5 的服务器上,为在gutub 上 ...
- ashx一般处理程序和HttpHandler
asp.net项目中,使用.ashx的文件(一般处理程序)可以用于处理客户端发送来的请求,并将服务器端的处理结果返回给客户端.它能返回的类型可以是文本.或者图片.有时候,我们可以在项目中使用.cs的文 ...
- C++ 静态static 变量在 cocos2d-x 里面使用误区
void Cms::showMonster(CCArray* monsterArray,int type) { <span style="color:#ff0000;"> ...