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. BZOJ 4034 [HAOI2015]T2(树链剖分)

    [题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=4034 [题目大意] 有一棵点数为 N 的树,以点 1 为根,且树点有边权. 有 M 个 ...

  2. (Android) Chinese Character

    Convert Chinese strings to English strings Apply pinyin4j.jar public static class ConvertChineseToPi ...

  3. Html 语法学习笔记二

    1.图像标签(<img>)和源属性(Src) 在 HTML 中.图像由 <img> 标签定义.        <img> 是空标签,意思是说,它仅仅包括属性,而且没 ...

  4. JavaScipt实现倒计时方法总结

    JavaScript中提供了两种实现计时.延时的方法,分别如下: 一. t = setTimeout(“function()", millisecond) 与 clearTimeout(t) ...

  5. android UI跨线程操作

    android应用程序也是单线程程序,主线程为UI线程. android 线程是非安全的,即不要在子线程中更新 UI. public class MasterActivity extends Acti ...

  6. 1.语言概述-JavaScript权威指南笔记

    上周三的时候交给老板目前的项目第一个迭代回顾会的总结.原本是以综述性的表述方式写的,交给他之后表示程序员不要长篇大论.总结要分为优点缺点期望等等块,每块列出条目,简明扼要的表达出来.这里也用这种风格. ...

  7. Vector(容器)

    vector(容器)就像数组一样,但比数组强大很多,下面介绍一下vector常用的几种方法: 一.对于vector自身的处理,包括赋初始值,复制等等: vector<int> v1 ; v ...

  8. PHP练习项目笔记之COOKIES

    主要是在登录和退出的时候,设置cookies.来保存登录和安全退出 1:在登录页面设置 //设置cookies的值 _setcookies($_rows['tg_username'], $_rows[ ...

  9. 利用python进行数据分析之pandas库的应用(二)

    本节介绍Series和DataFrame中的数据的基本手段 重新索引 pandas对象的一个重要方法就是reindex,作用是创建一个适应新索引的新对象 >>> from panda ...

  10. c# zip file and folder programmatically

    In .net 4.5 Framework, we can zip a file by this way: private static string CompressFile(string sour ...