POJ3186(KB12-O DP)

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5801		Accepted: 3003

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.

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

Sample Output

43

Hint

Explanation of the sample: 

Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2). 

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.

Source

USACO 2006 February Gold & Silver

 

//2017-04-06
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2005;
int v[N], n, dp[N][N];//dp[l][r]表示区间l~r间的最大收益
//状态转移方程：dp[l][r] = max(dp[l+1][r]+day*v[l], dp[l][r-1]+day*v[r])

int dfs(int l, int r, int day)
{
    if(l > r)return 0;
    if(dp[l][r])return dp[l][r];
    if(l == r)return dp[l][r] = day*v[l];
    return dp[l][r] = max(dfs(l+1, r, day+1)+day*v[l], dfs(l, r-1, day+1)+day*v[r]);
}

int main()
{
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 0; i < n; i++)
              scanf("%d", &v[i]);
        memset(dp, 0, sizeof(dp));
        int ans = dfs(0, n-1, 1);
        printf("%d\n", ans);
    }

    return 0;
}