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

--------------------------------------------------------------------------------------------

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
 
#define rep( i , n ) for( int i = 0 ; i < n ; i++ )
#define clr( x , c ) memset( x , c , sizeof( x ) )
 
using namespace std;
 
const int maxn = 2000 + 5;
 
int d[ maxn ][ maxn ];
int V[ maxn ];
int n;
 
int dp( int l , int r ) {
int &ans = d[ l ][ r ];
if( ans != -1 )
   return ans;
   
ans = max( dp( l + 1 , r ) + ( n - r + l ) * V[ l ] , dp( l , r - 1 ) + ( n - r + l ) * V[ r ] );
return ans;
}
int main() {
// freopen( "test.in" , "r" , stdin );
clr( d , -1 );
cin >> n;
rep( i , n ) {
   scanf( "%d" , V + i );
   
   d[ i ][ i ] = n * V[ i ];
   
}
cout << dp( 0 , n - 1 ) << "\n";
return 0;
}

--------------------------------------------------------------------------------------------

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:

•零食按照1..N编号,它们被排成一列放在一个很长的盒子里.盒子的两端都有开口,约翰每
  天可以从盒子的任一端取出最外面的一个.
•与美酒与好吃的奶酪相似,这些零食储存得越久就越好吃.当然,这样约翰就可以把它们卖出更高的价钱.
  •每份零食的初始价值不一定相同.约翰进货时,第i份零食的初始价值为Vi(1≤Vi≤1000).
  •第i份零食如果在被买进后的第a天出售,则它的售价是vi×a.
  Vi的是从盒子顶端往下的第i份零食的初始价值.约翰告诉了你所有零食的初始价值,并希望你能帮他计算一下,在这些零食全被卖出后,他最多能得到多少钱.

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

5
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

43

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

  1. 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 ...

  2. 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< ...

  3. 【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 ...

  4. [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 ...

  5. 【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+ ...

  6. 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[ ...

  7. BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚( 线段树 )

    线段树.. -------------------------------------------------------------------------------------- #includ ...

  8. 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   ...

  9. 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

  1. java 线程学习

    转载:详见处http://lavasoft.blog.51cto.com/62575/27069   Java多线程编程总结   下面是Java线程系列博文的一个编目:   Java线程:概念与原理 ...

  2. RTTI-CLASS

    package com.xt.test; interface Test1Interface { } interface Test2Interface { } class Test1 implement ...

  3. Optipng—PNG的优化图像工具初探

    PNG 即 Portable Network Graphic 的简称,PNG 图像是一种无损压缩图像文件格式.因为网络传输的需要,我们总是希望 PNG 图像的容量能够小些.小些.再小些.要优化 PNG ...

  4. 用DBMS_ADVISOR.SQLACCESS_ADVISOR创建SQL Access Advisor访问优化建议

    使用OEM方式来创建SQL Access Advisor访问优化建议,已经是四五年的事了,下面就来写写怎样使用DBMS_ADVISOR.SQLACCESS_ADVISOR来创建SQL Access A ...

  5. IOS学习之蓝牙4.0

    1建立中心角色 1 2 3 #import <CoreBluetooth/CoreBluetooth.h>  CBCentralManager *manager;  manager = [ ...

  6. linux下curl编程

    LibCurl是免费的客户端URL传输库,支持FTP,FTPS, HTTP, HTTPS, SCP, SFTP, TFTP, TELNET, DICT, FILE ,LDAP等协议,其主页是http: ...

  7. c 跟字符串有关的函数

    1.字符串查找 strstr char * strstr(const char *s1, const char *s2); 在s1中查找s2,如果找到返回首个s2的首地址 char * strcase ...

  8. 与html相关的知识点整理

    梳理html时发现的一些问题.总结一下,答案大都从网上找来. 一.html 与 htm 的区别 没有本质的区别..htm是在win32时代,系统只能识别3位扩展名时使用的.现在一般都使用.html. ...

  9. Android adb 命令图解

    做了这么长时间的开发与管理,在命令上总是自见则过,往往却忽视了在其命令上的分享过程,所以现在稍微有点时间就把 其命令的相关操作来简单的扫盲一番吧,也系统通过这种方式去授之以渔而不是鱼,好了,我以图解的 ...

  10. 【转载】设备坐标(窗口/window)和逻辑坐标(视口/viewport)

    一.预备知识 1.窗口是基于逻辑坐标的. 2.视口是基于设备坐标. 3.设备坐标是以像素为单位的,逻辑坐标是以.cm,m,mm,..... 4.系统最后一定要把逻辑坐标变为设备坐标. 5.设备坐标有3 ...