216 - 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 chained together 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 shown above, 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 at which 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.
题意:把所有电脑连成一条线,使得所用的电缆总长度最小。
两台电脑之间由一条缆线连接, 缆线的长度除了这两点间的直线长度,还要额外加上16米长。
一、暴力法
不同的方案就是不同的连接顺序,暴力枚举所有的方案,求出所有方案所用的电缆长度,求出一种长度最短的方案。设每台电脑的编号依次为0,1,2,……,n-1,根据连接顺序,利用STL里面的next_permutation函数求出0~n-1的全排列,每一种排列对应一种连接方案。因为最大只有8台电脑,所以数据量比较小,可以实现。
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct cable
{
double x,y;
}a[10]; /*保存坐标*/
int b[10],c[10];
double length(double x1,double y1,double x2,double y2) /*求两点之间距离*/
{
double L=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
return L;
}
int main()
{
double sum,min,l;
int n,i,cases=0;
while(~scanf("%d",&n)&&n)
{
for(i=0;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
b[i]=i;
}
min=2147483645;
do
{
sum=0;
for(i=0;i<n-1;i++)
{
l=length(a[b[i]].x,a[b[i]].y,a[b[i+1]].x,a[b[i+1]].y)+16;
sum+=l;
}
if(sum<min)
{
min=sum;
for(i=0;i<n;i++)
c[i]=b[i]; //可用 memcpy(c,b,sizeof(b)) 代替
}
}while(next_permutation(b,b+n));
printf("**********************************************************\n");
printf("Network #%d\n",++cases);
for(i=0;i<n-1;i++)
{
l=length(a[c[i]].x,a[c[i]].y,a[c[i+1]].x,a[c[i+1]].y)+16;
printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",a[c[i]].x,a[c[i]].y,a[c[i+1]].x,a[c[i+1]].y,l);
}
printf("Number of feet of cable required is %.2lf.\n",min);
}
return 0;
}

因为next_permutation是C++里面的函数,所以提交时要选择C++语言,不然会编译错误。

二、回溯法
暴力很容易想到,但是不太灵活,也并不是所有都适用的。

而回溯法是更常用的方法,也更加灵活,更难掌握。

回溯法就是深搜(DFS)的变形。 一般深搜是要访问所有的解答树的,而回溯也是把问题分成若干步骤并递归求解,但是如果当前步骤已经不是最佳选择的

话,就不继续递归下去,而是返回上一及的递归调用。这样就可以节省很多的时间,而不必徒劳去访问那些“不归路”。

#include<stdio.h>
#include<string.h>
#include<math.h>
struct cable
{
double x,y;
}a[10];
int b[10],c[10],vis[10],n;
double min,sum,l;
double length(double x1,double y1,double x2,double y2)
{
double L=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2))+16;
return L;
}
void dfs(int cur,double sum)
{
int i;
if(cur==n)
{
if(sum<min)
{
min=sum;
memcpy(c,b,sizeof(b));
}
return;
}
if(sum>=min) return;
for(i=0;i<n;i++)
{
if(vis[i]) continue;
vis[i]=1;
b[cur]=i;
if(cur==0)
dfs(cur+1,0);
else
{
l=length(a[b[cur]].x,a[b[cur]].y,a[b[cur-1]].x,a[b[cur-1]].y);
dfs(cur+1,sum+l);
}
vis[i]=0;
}
}
int main()
{
int cases=0,i;
while(~scanf("%d",&n)&&n)
{
memset(vis,0,sizeof(vis));
for(i=0;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
b[i]=i;
}
min=99999999;
dfs(0,0);
printf("**********************************************************\n");
printf("Network #%d\n",++cases);
for(i=1;i<n;i++)
{
l=length(a[c[i-1]].x,a[c[i-1]].y,a[c[i]].x,a[c[i]].y);
printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",a[c[i-1]].x,a[c[i-1]].y,a[c[i]].x,a[c[i]].y,l);
}
printf("Number of feet of cable required is %.2lf.\n",min);
}
return 0;
}

UVA 216 - Getting in Line的更多相关文章

  1. uva 216 Getting in Line 最短路,全排列暴力做法

    题目给出离散的点,要求求出一笔把所有点都连上的最短路径. 最多才8个点,果断用暴力求. 用next_permutation举出全排列,计算出路程,记录最短路径. 这题也可以用dfs回溯暴力,但是用最小 ...

  2. UVa 216 Getting in Line【枚举排列】

    题意:给出n个点的坐标,(2<=n<=8),现在要使得这n个点连通,问最小的距离的和 因为n很小,所以可以直接枚举这n个数的排列,算每一个排列的距离的和, 保留下距离和最小的那个排列就可以 ...

  3. Getting in Line UVA 216

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

  4. UVA题目分类

    题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...

  5. codeforces 713B B. Searching Rectangles(二分)

    题目链接: B. Searching Rectangles time limit per test 1 second memory limit per test 256 megabytes input ...

  6. Python写出LSTM-RNN的代码

    0. 前言 本文翻译自博客: iamtrask.github.io ,这次翻译已经获得trask本人的同意与支持,在此特别感谢trask.本文属于作者一边学习一边翻译的作品,所以在用词.理论方面难免会 ...

  7. Codeforces Round #371 (Div. 2) D. Searching Rectangles 交互题 二分

    D. Searching Rectangles 题目连接: http://codeforces.com/contest/714/problem/D Description Filya just lea ...

  8. hdu 5735 Born Slippy 暴力

    Born Slippy 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5735 Description Professor Zhang has a r ...

  9. 转:西部数据NAS设备hack

    通过该文学习一下常见硬件web漏洞.重点关注一下几个方面: 1.登录验证代码: 2.文件上传代码: 3.system/exec/popen等是否存在注入可能: 4.调用二进制文件: 5.未登陆可以访问 ...

随机推荐

  1. SQL语句分享[不定期更新]

    查询临时表 if object_id('')>0 查询表中的数据 select 'insert into ta1(col1,col2,col3) values('''+ltrim(列1)+''' ...

  2. 分布式基础学习(2)分布式计算系统(Map/Reduce)

    二. 分布式计算(Map/Reduce) 分 布式式计算,同样是一个宽泛的概念,在这里,它狭义的指代,按Google Map/Reduce框架所设计的分布式框架.在Hadoop中,分布式文件 系统,很 ...

  3. 接口自动化测试:Thrift框架RPC协议客户端开发

    import java.lang.Thread.State;import java.util.Iterator;import java.util.List; import org.apache.thr ...

  4. WebApp 框架

    Razor 在WebApp 框架的运用   前面有两章介绍了WebApp框架<WebApp MVC,“不一样”的轻量级互联网应用程序开发框架>和<WebApp MVC 框架的开发细节 ...

  5. 快速构建Windows 8风格应用22-MessageDialog

    原文:快速构建Windows 8风格应用22-MessageDialog 本篇博文主要介绍MessageDialog概述.MessageDialog常用属性和方法.如何构建MessageDialog ...

  6. windows下oracle数据库定时备份与压缩批处理脚本(win7/win2008亲测通过)

    第一种方式:全备份,并直接覆盖原来的备份文件,固定命名. 脚本: @echo off exp system/a123@HZPG file=e:\db_backup\oradata.dmp log=e: ...

  7. [译]ava 设计模式之构造器

    (文章翻译自Java Design Pattern: Builder) 构造器模式的关键之处在于它使用一步接招一步的流程去构建东西,例如:尽管构建的每一步是不相同的但是每一个产品还是遵循相同的流程. ...

  8. 删除sql server中重复的数据

    原文:删除sql server中重复的数据 with list_numbers as( select Name, AuthorOrTime, Url, Price, EstimatePrice, Si ...

  9. 【转】【Android应用开发详解】第01期:第三方授权认证(一)实现第三方授权登录、分享以及获取用户资料

    转载请注明出处:http://blog.csdn.net/yangyu20121224/article/details/9057257 由于公司项目的需要,要实现在项目中使用第三方授权登录以及分享文字 ...

  10. Delphi的注册表操作

    转帖:Delphi的注册表操作 2009-12-21 11:12:52 分类: Delphi的注册表操作 32位Delphi程序中可利用TRegistry对象来存取注册表文件中的信息.     一.创 ...