• 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

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

    #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 ) << " ";
    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

  • 相关阅读:
    Python 容器用法整理
    C/C++中浮点数格式学习——以IEEE75432位单精度为例
    关于C/C++中的位运算技巧
    [GeekBand] C++11~14
    [GeekBand] 探讨C++新标准之新语法——C++ 11~14
    [GeekBand] 面向对象的设计模式(C++)(2)
    [GeekBand] 面向对象的设计模式(C++)(1)
    [GeekBand] STL与泛型编程(3)
    [GeekBand] STL与泛型编程(2)
    [GeekBand] STL与泛型编程(1)
  • 原文地址:https://www.cnblogs.com/JSZX11556/p/4558177.html
Copyright © 2020-2023  润新知