Problem W UVA 662 二十三 Fast Food
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
System Crawler (2015-08-27)
Description
The fastfood chain McBurger owns several restaurants along a highway. Recently, they have decided to build several depots along the highway, each one located at a restaurent and supplying several of the restaurants with the needed ingredients. Naturally, these depots should be placed so that the average distance between a restaurant and its assigned depot is minimized. You are to write a program that computes the optimal positions and assignments of the depots.
To make this more precise, the management of McBurger has issued the following specification: You will be given the positions of nrestaurants along the highway as n integers (these are the distances measured from the company's headquarter, which happens to be at the same highway). Furthermore, a number
will be given, the number of depots to be built.
The k depots will be built at the locations of k different restaurants. Each restaurant will be assigned to the closest depot, from which it will then receive its supplies. To minimize shipping costs, the total distance sum, defined as
must be as small as possible.
Write a program that computes the positions of the k depots, such that the total distance sum is minimized.
Input
The input file contains several descriptions of fastfood chains. Each description starts with a line containing the two integers n and k. nand k will satisfy ,
,
. Following this will n lines containing one integer each, giving the positions di of the restaurants, ordered increasingly.
The input file will end with a case starting with n = k = 0. This case should not be processed.
Output
For each chain, first output the number of the chain. Then output an optimal placement of the depots as follows: for each depot output a line containing its position and the range of restaurants it serves. If there is more than one optimal solution, output any of them. After the depot descriptions output a line containing the total distance sum, as defined in the problem text.
Output a blank line after each test case.
Sample Input
6 3
5
6
12
19
20
27
0 0
Sample Output
Chain 1
Depot 1 at restaurant 2 serves restaurants 1 to 3
Depot 2 at restaurant 4 serves restaurants 4 to 5
Depot 3 at restaurant 6 serves restaurant 6
Total distance sum = 8
Miguel Revilla
2000-05-22
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int inf=0x3f3f3f3f; int a[],face[][],dp[][],cost[][]; int put(int k,int n)
{
if(k)
{
int up=face[k][n]+,mid=(up+n)/;
put(k-,face[k][n]);
if(up==n)
printf("Depot %d at restaurant %d serves restaurant %d\n",k,mid,n);
else
printf("Depot %d at restaurant %d serves restaurants %d to %d\n",k,mid,up,n);
}
return ;
} int main()
{
int n,K,mid,cas=;
int i,j,k;
while(scanf("%d %d",&n,&K)!=EOF)
{
memset(cost,,sizeof(cost));
if(n== && K==)
break;
for(i=;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(i=;i<=n;i++)
{
for(j=i;j<=n;j++)
{
mid=(i+j)/;
for(k=i;k<=j;k++)
{
cost[i][j]=cost[i][j]+abs(a[k]-a[mid]);
}
}
}
for(i=;i<=n;i++)
{
dp[][i]=cost[][i];
} for(i=;i<=K;i++)
{
for(j=i;j<=n;j++)
{
dp[i][j]=inf;
for(k=i-;k<j;k++)
{
if(dp[i][j]>(dp[i-][k]+cost[k+][j]))
dp[i][j]=dp[i-][k]+cost[k+][j],face[i][j]=k;
}
}
} printf("Chain %d\n",cas++);
put(K,n);
printf("Total distance sum = %d\n\n",dp[K][n]);
}
return ;
}
Problem W UVA 662 二十三 Fast Food的更多相关文章
- 【Visual C++】游戏开发五十六 浅墨DirectX教程二十三 打造游戏GUI界面(一)
本系列文章由zhmxy555(毛星云)编写,转载请注明出处. 文章链接:http://blog.csdn.net/poem_qianmo/article/details/16384009 作者:毛星云 ...
- 木块问题(The Blocks Problem,Uva 101)
不定长数组:vector vector就是一个不定长数组.不仅如此,它把一些常用操作“封装”在了vector类型内部. 例如,若a是一个vector,可以用a.size( )读取它的大小,a.resi ...
- WPF入门教程系列二十三——DataGrid示例(三)
DataGrid的选择模式 默认情况下,DataGrid 的选择模式为“全行选择”,并且可以同时选择多行(如下图所示),我们可以通过SelectionMode 和SelectionUnit 属性来修改 ...
- Bootstrap <基础二十三>页面标题(Page Header)
页面标题(Page Header)是个不错的功能,它会在网页标题四周添加适当的间距.当一个网页中有多个标题且每个标题之间需要添加一定的间距时,页面标题这个功能就显得特别有用.如需使用页面标题(Page ...
- Web 前端开发精华文章推荐(HTML5、CSS3、jQuery)【系列二十三】
<Web 前端开发精华文章推荐>2014年第2期(总第23期)和大家见面了.梦想天空博客关注 前端开发 技术,分享各类能够提升网站用户体验的优秀 jQuery 插件,展示前沿的 HTML5 ...
- iOS安全攻防(二十三):Objective-C代码混淆
iOS安全攻防(二十三):Objective-C代码混淆 class-dump能够非常方便的导出程序头文件,不仅让攻击者了解了程序结构方便逆向,还让着急赶进度时写出的欠完好的程序给同行留下笑柄. 所以 ...
- WCF技术剖析之二十三:服务实例(Service Instance)生命周期如何控制[下篇]
原文:WCF技术剖析之二十三:服务实例(Service Instance)生命周期如何控制[下篇] 在[第2篇]中,我们深入剖析了单调(PerCall)模式下WCF对服务实例生命周期的控制,现在我们来 ...
- Bootstrap入门(二十三)JS插件1:模态框
Bootstrap入门(二十三)JS插件1:模态框 1.静态实例 2.动态实例 3.模态框的尺寸和效果 4.包含表单的模态框 模态框经过了优化,更加灵活,以弹出对话框的形式出现,具有最小和最实用的功能 ...
- JAVA之旅(二十三)——System,RunTime,Date,Calendar,Math的数学运算
JAVA之旅(二十三)--System,RunTime,Date,Calendar,Math的数学运算 map实在是太难写了,整理得我都晕都转向了,以后看来需要开一个专题来讲这个了,现在我们来时来学习 ...
随机推荐
- HGE引擎之hgeSprite
一.hgeSprite类 hgeSprite是一个精灵实体的HGE帮助类. 1.构造函数 创建和初始化一个hgeSprite对象. hgeSprite(HTEXTURE tex, float x, f ...
- SQL server 创建 修改表格 及表格基本增删改查 及 高级查询 及 (数学、字符串、日期时间)函数[转]
SQL server 创建 修改表格 及表格基本增删改查 及 高级查询 及 (数学.字符串.日期时间)函数 --创建表格 create table aa ( UserName varchar(50 ...
- Calibrating delay loop... 问题以及解决方法(RealARM开发板)
RealARM的210开发板在启动是有时会出现这样的死循环 Calibrating delay loop... ,那么原因是什么呢? 经过查找,发现跟RTC有关,实际上就是晶振和RTC电源的问题.所以 ...
- React笔记_(2)_react语法1
这一节内容主要以了解为主. 渐渐的体会react的语法和其特性. ① htmlAndJs 混合编写 react和以往的前后台书写方式不一样. 在之前的多个语言中,讲求的是将页面代码和js代码逻辑分开, ...
- 启动管理软件服务器时,提示midas.dll错误
首先确认系统以及管理软件目录内是否有midas.dll文件,如果没有,请复制或下载midas.dll到相应目录.系统默认路径为:'c:\windows\system32\' 然后依次打开“开始菜单”内 ...
- Google 开发新的开源系统 Fuchsia
google 最新os 下载 https://github.com/fuchsia-mirror/magenta 本文转自:http://www.oschina.net/news/76094/goog ...
- Valid Anagram
class Solution { public: bool isAnagram(string s, string t) { if(t=="") return s=="&q ...
- java IO和NIO 的区别
Java NIO和IO的主要区别 下表总结了Java NIO和IO之间的主要差别. IO NIO 面向流 面向缓冲 阻塞IO 非 ...
- RabbitMQ 基本概念介绍-----转载
1. 介绍 RabbitMQ是一个由erlang开发的基于AMQP(Advanced Message Queue )协议的开源实现.用于在分布式系统中存储转发消息,在易用性.扩展性.高可用性等方面都非 ...
- java.lang.IllegalThreadStateException
java.lang.IllegalThreadStateException 今天遇到了这个问题.当时的情景是想要循环实现了runable的类和继承Thread类的两个线程.可是没有注意到,继承自Thr ...