[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