# BZOJ 1652: [Usaco2006 Feb]Treats for the Cows (区间dp)

Posted by 111qqz on Tuesday, April 12, 2016

## 1652: [Usaco2006 Feb]Treats for the Cows

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 290  Solved: 226 [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.

•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.

dp[i][j]表示最左端为i，最右端为j能得到的最大价值。

**那么t=n+i-j,**j = n-k+1;

*转移方程为dp[i][j] = max(dp[i+1][j]+t*v[i],dp[i][j-1]v[j])

（注意观察，（i,j）的状态是由(i+1,j)或者(i,j-1)得到的，因为是从里向外推。）

``````/* ***********************************************
Author :111qqz
Created Time :2016年04月11日 星期一 20时46分42秒
File Name :code/bzoj/1652.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=2E3+3;
int dp[N][N];//dp[i][j]表示最左边取i,最右边取j的最大值。
int n;
int v[N];
bool vis[N];
int main()
{
#ifndef  ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif

scanf("%d",&n);
v[0] = 0 ;
for ( int i = 1 ; i <= n ; i++) scanf("%d",&v[i]);
ms(dp,0);

for ( int k = 1 ; k <= n ; k++) //k是i两段的间隔距离，也可以理解成剩余的长度。k=j-i+1;
{				//这个划分状态的技巧有点厉害
for ( int i = 1 ; i+k-1 <= n  ; i++)
{
int t = n-k+1;
int j = i+k-1;
//		cout<<"i:"<<i<<" j:"<<j<<" t:"<<t<<endl;
dp[i][j] = max(dp[i+1][j]+t*v[i],dp[i][j-1]+t*v[j]); //转移是倒着进行的，从内向外
}
}

printf("%d\n",dp[1][n]);

#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
``````