Treats for the Cows (区间DP)

Treats for the Cows

 POJ - 3186 

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.

题意：

　　给一个数组v，每次可以取前面的或者后面的，第k次取的v[i]价值为v[i]*k，问总价值最大是多少。

题解：

　　区间DP

　　设dp[i][j]为从i取到j的最大值，假设每个数都是最后一个被取，那么dp[i][i]=a[i]*n。而dp[i][j]只能由dp[i+1][j]和dp[i][j-1]转移来，所以状态转移方程为dp[l][r] = Math.max(dp[l][r], Math.max(dp[l+1][r]+a[l]*(n-len), dp[l][r-1]+a[r]*(n-len)));其中（n-len）表示第几个取。

import java.util.Scanner;

public class Main { 
    static int casen,n;
    static int [][] dp ;
    static int [] a = new int [2019];
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        n = cin.nextInt();
        dp = new int [n+10][n+10];
        for(int i=1;i<=n;i++) {
            a[i] = cin.nextInt();
            dp[i][i] = a[i]*n;
        }
        long ans = 0;
        for(int len=1;len<=n;len++) {
            for(int l=1;l+len<=n;l++) {
                int r = l+len;
                dp[l][r] = Math.max(dp[l][r], Math.max(dp[l+1][r]+a[l]*(n-len), dp[l][r-1]+a[r]*(n-len)));
            }
        }
        System.out.println(dp[1][n]);
    }
}