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

  1. 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: ...

  2. 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 ...

  3. UVA 216 - Getting in Line

    216 - Getting in Line Computer networking requires that the computers in the network be linked. This ...

  4. Getting in Line UVA 216

     Getting in Line  Computer networking requires that the computers in the network be linked. This pro ...

  5. 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 & ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. chrome使用技巧(转)good

    阅读目录 写在前面 快速切换文件 在源代码中搜索 在源代码中快速跳转到指定的行 使用多个插入符进行选择 设备模式 设备传感仿真 格式化凌乱的js源码 颜色选择器 改变颜色格式 强制改变元素状态(方便查 ...

  2. chrome主页被篡改为360该溶液的导航

    昨天,安装游戏后,,发现chrome该主页被篡改为360导航. 进入chrome设置更改主页,再次启动chrome或360导航,后来头发今天chrome快捷方式目标再加上一堆的属性后面360网站导航, ...

  3. Apple Watch 1.0 开发介绍 1.2 简介 配置Xcode工程

    WatchKit app需要一个现有的iOS app.在iOS app工程中,添加一个新的WatchKit app target,它包含了WatchKit app和WatchKit extension ...

  4. 【Android基础】Activity之间进行参数传递的三种方式

    1.使用Intent进行传输 //发送数据的Activity class button implements OnClickListener{ @Override public void onClic ...

  5. 怎样用C#代码管理SharePoint解决方式

    怎样用C#代码管理SharePoint解决方式         本文我们将了解怎样用代码管理SharePoint解决方式.我们使用server端对象模型抽取解决方式.         SharePoi ...

  6. 猫学习IOS(四)UI半小时就搞定Tom猫

    阿土 首先对影响 下载项目的源材料: Tom猫游戏代码iOS 素材http://blog.csdn.net/u013357243/article/details/44457357 效果图 以前风靡一时 ...

  7. overflow的几个坑

    在android 4.0的原生浏览器上注意: html元素上不要加overflow: auto;的样式,否则会造成有些元素无法点击 在absolute元素上 不要加 overflow: auto; 否 ...

  8. Thread thread2 = new Thread()

    Thread thread2 = new Thread() { @Override public void run() { test.function(); } }; thread1.start(); ...

  9. C#基础之二

    C#基础之二 1.  静态字段.静态成员:由该类的所有实例共享,其变化,实例是对所有可见. 2.  声明为static有:场.方法.属性.构造函数.运营商.事件. 3.  不同的常量静态量:位置,在编 ...

  10. javascript 模块化编程 1

    var myModule=(function(){ var models={}; function define(name,deps,cb){ var tk=[]; for(var i=0,l=dep ...