BZOJ 1652: [Usaco2006 Feb]Treats for the Cows( dp )

dp( L , R ) = max( dp( L + 1 , R ) + V_L * ( n - R + L ) , dp( L , R - 1 ) + V_R * ( n - R + L ) )
边界 : dp( i , i ) = V[ i ] * n
--------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------
1652: [Usaco2006 Feb]Treats for the Cows
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 250 Solved: 199
[Submit][Status][Discuss]
Description
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time. The treats are interesting for many reasons: * The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats. * Like fine wines and delicious cheeses, the treats improve with age and command greater prices. * The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000). * Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a. Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally? The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
约翰经常给产奶量高的奶牛发特殊津贴,于是很快奶牛们拥有了大笔不知该怎么花的钱.为此,约翰购置了N(1≤N≤2000)份美味的零食来卖给奶牛们.每天约翰售出一份零食.当然约翰希望这些零食全部售出后能得到最大的收益.这些零食有以下这些有趣的特性:
Input
* Line 1: A single integer,
N * Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
* Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input
1
3
1
5
2
Five treats. On the first day FJ can sell either treat #1 (value 1) or
treat #5 (value 2).
Sample Output
OUTPUT DETAILS:
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order
of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
HINT
Source
BZOJ 1652: [Usaco2006 Feb]Treats for the Cows( dp )的更多相关文章
- BZOJ 1652: [Usaco2006 Feb]Treats for the Cows
题目 1652: [Usaco2006 Feb]Treats for the Cows Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 234 Solve ...
- bzoj 1652: [Usaco2006 Feb]Treats for the Cows【区间dp】
裸的区间dp,设f[i][j]为区间(i,j)的答案,转移是f[i][j]=max(f[i+1][j]+a[i](n-j+i),f[i][j-1]+a[j]*(n-j+i)); #include< ...
- 【BZOJ】1652: [Usaco2006 Feb]Treats for the Cows(dp)
http://www.lydsy.com/JudgeOnline/problem.php?id=1652 dp.. 我们按间隔的时间分状态k,分别为1-n天 那么每对间隔为k的i和j.而我们假设i或者 ...
- [BZOJ 1652][USACO 06FEB]Treats for the Cows 题解(区间DP)
[BZOJ 1652][USACO 06FEB]Treats for the Cows Description FJ has purchased N (1 <= N <= 2000) yu ...
- 【记忆化搜索】bzoj1652 [Usaco2006 Feb]Treats for the Cows
跟某NOIP的<矩阵取数游戏>很像. f(i,j)表示从左边取i个,从右边取j个的答案. f[x][y]=max(dp(x-1,y)+a[x]*(x+y),dp(x,y-1)+a[n-y+ ...
- BZOJ1652 [Usaco2006 Feb]Treats for the Cows
蒟蒻许久没做题了,然后连动规方程都写不出了. 参照iwtwiioi大神,这样表示区间貌似更方便. 令f[i, j]表示i到j还没卖出去,则 f[i, j] = max(f[i + 1, j] + v[ ...
- BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚( 线段树 )
线段树.. -------------------------------------------------------------------------------------- #includ ...
- BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚
题目 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 553 ...
- poj 3186 Treats for the Cows(dp)
Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for gi ...
随机推荐
- Linux Apache绑定多域名
1 网上查到资源不符 网上查到的Apache绑定域名都说要修改http.conf文件,但是我的服务器上的apache是通过apt-get install安装的,安装方法应该是没错的,但是通过find ...
- 类比的方法学习Performance_schema
引用自:http://www.javacoder.cn/?p=332 MySQL在5.6版本中包含了一个强大的特性——performance-schema,合理的使用这个数据库中的表,能为我们解决一些 ...
- Linux Kernel 2.6.28 以上有BUG,系统运行第208.5天down机
简介: 业务服务器有一台服务器出现意外down机,服务器ping 不通.无法登陆,本想通过公司KVM系统登陆系统重启解决,登陆KVM后发现系统屏幕打印大量的内核错误,KVM无法使用.无法发送重启服务器 ...
- Qt项目管理(33个规则)
2016-06-20 花莫弦 小小杂货铺LY 一.qmake的介绍 qmake是Trolltech公司创建的用来为不同的平台和编译器书写Makefile的工具. 手写Makefile是比较困难并且容易 ...
- kafka初探
http://www.infoq.com/cn/articles/kafka-analysis-part-1
- easyui-layout中的收缩层无法显示标题问题解决
先看问题描述效果图片: 如上,我的查询条件是放在layout下面的一个可收缩层中,初始是收缩的,title显示不出来的话对使用者很不方便,代码如下: <div id="__MODULE ...
- 使用vi编辑binary文件
原理:使用xxd将当前文件转成hex格式,编辑,然后再转回去 /usr/bin/xxd xxd - make a hexdump or do the reverse 例子: 用binary模式启动vi ...
- php获取apk信息
使用方法如下: <?php require('apk_parser.php'); $p = new ApkParser(); /* if($argc<2) { echo "usa ...
- 基于Proxy思想的Android插件框架
意义 研究插件框架的意义在于下面几点: 减小安装包的体积,通过网络选择性地进行插件下发 模块化升级.减小网络流量 静默升级,用户无感知情况下进行升级 解决低版本号机型方法数超限导致无法安装的问题 代码 ...
- ceph基本操作整理
一.ceph基本操作: 启动osd.mon进程: start ceph-osd id=X start ceph-mon id=YYY 关闭osd.mon进程: stop ceph-osd id=X ...