[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) 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

Solution

1.初始化

f[i][i]代表这个只有这个物品卖出的利润,显然此时f[i][i]=v[i],同时记录v[i]的前缀和,用于转移。

2.DP

方程为f[l][r]=max(f[l+1][r],f[l][r-1])+v[r]-v[l-1],答案由两种小1长度的区间得到,加上区间和代表所有区间内的物品都延迟一天卖出。

Code

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define R register
using namespace std;

int a[2010],v[2010],f[2010][2010];

inline int rd(){
    int x=0;
    bool f=1;
    char c=getchar();
    while(!isdigit(c)){
        if(c=='-') f=0;
        c=getchar();
    }
    while(isdigit(c)){
        x=(x<<1)+(x<<3)+(c^48);
        c=getchar();
    }
    return f?x:-x;
}

int main(){
    int n=rd();
    for(R int i=1;i<=n;++i){
        a[i]=f[i][i]=rd();
        v[i]=v[i-1]+a[i];
    }
    for(R int len=2;len<=n;++len)
        for(R int l=1;l<=n-len+1;++l){
            int r=l+len-1;
            f[l][r]=max(f[l+1][r],f[l][r-1])+v[r]-v[l-1]; 
        }
    printf("%d",f[1][n]);
    return 0;
}

有关区间DP的其他讲解参考我的随笔:http://www.cnblogs.com/COLIN-LIGHTNING/p/9038198.html

posted @ 2018-05-15 19:19 COLINGAO 阅读(...) 评论(...) 编辑 收藏