As you may know from the comic \Asterix and the Chieftain's Shield", Gergovia consists of one street,
and every inhabitant of the city is a wine salesman. You wonder how this economy works? Simple
enough: everyone buys wine from other inhabitants of the city. Every day each inhabitant decides how
much wine he wants to buy or sell. Interestingly, demand and supply is always the same, so that each
inhabitant gets what he wants.
There is one problem, however: Transporting wine from one house to another results in work. Since
all wines are equally good, the inhabitants of Gergovia don't care which persons they are doing trade
with, they are only interested in selling or buying a specic amount of wine. They are clever enough
to gure out a way of trading so that the overall amount of work needed for transports is minimized.
In this problem you are asked to reconstruct the trading during one day in Gergovia. For simplicity
we will assume that the houses are built along a straight line with equal distance between adjacent
houses. Transporting one bottle of wine from one house to an adjacent house results in one unit of
work.
Input
The input consists of several test cases. Each test case starts with the number of inhabitants n
(2 n 100000). The following line contains n integers ai (

uva 11054 wine trading in gergovia (归纳【好吧这是我自己起的名字】)——yhx的更多相关文章

  1. UVA 11054 Wine trading in Gergovia 葡萄酒交易 贪心+模拟

    题意:一题街道上很多酒店,交易葡萄酒,正数为卖出葡萄酒,负数为需要葡萄酒,总需求量和总售出量是相等的,从一家店到另外一家店需要路费(路费=距离×运算量),假设每家店线性排列且相邻两店之间距离都是1,求 ...

  2. UVa 11054 Wine trading in Gergovia【贪心】

    题意:给出n个等距离的村庄,每个村庄要么买酒,要么卖酒,买酒和卖酒的总量相等, 把k个单位的酒从一个村庄运送到相邻的村庄,需要耗费k个单位劳动力,问怎样运送酒使得耗费的劳动力最少 买     卖    ...

  3. UVa 11054 Wine trading in Gergovia

    题意: 直线上有n个等距的酒庄,每个酒庄对酒的需求量为ai(正数说明需要买酒,负数需要卖酒),而且保证所有的酒庄供需平衡. 搬运x个单位的酒到相邻的酒庄需要x个劳动力,求要使所有酒庄供需平衡最少需要多 ...

  4. UVA 11054 Wine trading in Gergovia(思维)

    题目链接: https://vjudge.net/problem/UVA-11054 /* 问题 输入村庄的个数n(2=<n<=100000)和n个村庄的数值,正代表买酒,负代表卖酒,k个 ...

  5. UVA - 11054 Wine trading in Gergovia 扫描法

    题目:点击打开题目链接 思路:考虑第一个村庄,如果第一个村庄需要买酒,那么a1>0,那么一定有劳动力从第二个村庄向第一个村庄搬酒,这些酒可能是第二个村庄生产的,也可能是从其他村庄搬过来的,但这一 ...

  6. UVA - 11054 Wine trading in Gergovia (Gergovia 的酒交易)(贪心+模拟)

    题意:直线上有n(2<=n<=100000)个等距的村庄,每个村庄要么买酒,要么卖酒.设第i个村庄对酒的需求为ai(-1000<=ai<=1000),其中ai>0表示买酒 ...

  7. 【例题 8-5 UVA - 11054】Wine trading in Gergovia

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 考虑第一个店. 如果它要酒的话,只能从第二个店那里运过来. 这样,问题就转化成后n-1个店的问题了. 然后会发现,第二家店它的情况也 ...

  8. 【uva 11054】Wine trading in Gergovia(算法效率--等价转换)

    题意:N个等距村庄,买(>0)卖(<0)酒,供需平衡,运K则需K劳动力.问所需的最小劳动力. 解法:由于运出或运入1的都需经过2,所以无论如何,都可以等价于从2本身运入或运出.因此可以将1 ...

  9. 8-5 Wine trading in Gergovia Gergovia的酒交易 uva11054

    等价转换思维题 题意: 直线上有n(2<=n<=100000)个等距的村庄  每个村庄要么买酒 要么卖酒  设第i个村庄对酒的需求量为ai  绝对值小于一千  其中ai大于0表示买酒   ...

随机推荐

  1. 【C#进阶系列】05 基元类型、引用类型和值类型

     基元类型和FCL类型 FCL类型就是指Int32这种类型,这是CLR支持的类型. 而基元类型就是指int这种类型,这是C#编译器支持的,实际上在编译后,还是会被转为Int32类型. 而且学过C的朋友 ...

  2. Genymotion Shell命令

    命令行选项 •-h 输出帮助信息 •-r ip_address 连接到特定Genymotion虚拟设备 •-c "command" 在原生shell环境中执行给定命令而后返回 •- ...

  3. 控制台(Console)报错:java.io.IOException: Broken pipe

    控制台(Console)输出: java.io.IOException: Broken pipe at sun.nio.ch.FileDispatcherImpl.write0(Native Meth ...

  4. 使用WebMatrix发布网站

    使用WebMatrix发布网站 WebMatrix 简介: Microsoft WebMatrix 是微软最新的 Web 开发工具,它包含了构建网站所需要的一切元素.您可以从开源 Web 项目或者内置 ...

  5. SharePoint 新特性及安装需知

    以下内容转自Kaneboy 大牛,但我在安装正式版的过程中发现一些问题,主要是.net 版本的问题,弄了我一个晚上,我在下面标出来了.我的安装环境是Windows server 2012 R2 关于详 ...

  6. PowerDesign设置列名与表名的大小写规则

    因为设计的Model是要用到Oracle 11g的,Oracle 11g的表名和列名默认是大写,在设计模型当中小写转大写一直很苦闹,发现PowerDedign可以直接设置,见下文: 打开Tools-- ...

  7. 转:纠结的Shim

    原文地址:http://www.haorooms.com/post/requirejs_sy_lj RequireJs已经流行很久了,它提供了以下功能: 声明不同js文件之间的依赖 可以按需.并行.延 ...

  8. 利用在线工具自动化生成findviewById

    我们在编码的时候经常会用到findviewById,不厌其烦,我之前介绍过一个很取巧的方法,挺好用的,这里再贴一下: public class KaleBaseActivity extends Act ...

  9. GCD中的dispatch_set_target_queue的用法及作用

    (一),使用dispatch_set_target_queue更改Dispatch Queue的执行优先级 dispatch_queue_create函数生成的DisPatch Queue不管是Ser ...

  10. block做方法参数时--block的参数传值过程 例1

    说明:此例子中方法的调用在此文中是从下到上调用的.(即:     方法五调用方法四:      方法四调用方法三) 方法一:- (void)setCompletionBlockWithSuccess: ...