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
- 生成输出url
继续使用前面的例子11-3URLTestDemo,修改Global.asax中的RegisterRoutes方法如下: public static void RegisterRoutes(RouteC ...
- 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 ...
- POJ 3723 Conscription(并查集建模)
[题目链接] http://poj.org/problem?id=3723 [题目大意] 招募名单上有n个男生和m个女生,招募价格均为10000, 但是某些男女之间存在好感,则招募的时候, 可以降低与 ...
- HDU 5765 Bonds(状压DP)
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5765 [题目大意] 给出一张图,求每条边在所有边割集中出现的次数. [题解] 利用状压DP,计算不 ...
- Ubuntu 麒麟版下安装:Apache+php5+mysql+phpmyadmin.
摘要 LAMP是Linux web服务器组合套装的缩写,分别是Apache+MySQL+PHP.此文记录在Ubuntu上安装Apache2服务器,包括PHP5(mod_php)+MySQL+phpmy ...
- fastDFS同步问题讨论
一.文件同步延迟问题 前面也讲过fastDFS同组内storage server数据是同步的, Storage server中由专门的线程根据binlog进行文件同步.为了最大程度地避免相互影响以及出 ...
- 套接字socket 的地址族和类型、工作原理、创建过程
注:本分类下文章大多整理自<深入分析linux内核源代码>一书,另有参考其他一些资料如<linux内核完全剖析>.<linux c 编程一站式学习>等,只是为了更好 ...
- js兼容性大全
js有个第二定律好的属性/选择器一定不兼容/* 获取类名通用代码*/function getClassName(){ if(document.getElementsByClassName){ doso ...
- java思考题
1.仔细阅读示例: EnumTest.java,运行它,分析运行结果? 你能得到什么结论?你掌握了枚举类型的基本用法了吗? public class EnumTest { public static ...
- Android 调用webservice faultactor 错误
1.错误:02-05 09:56:17.266: E/WebServiceUtil(801): --- 内部异常堆栈跟踪的结尾 ---' faultactor: 'null' detail: org. ...