POJ3186：Treats for the Cows(区间DP)

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

 

题意：给出的一系列的数字，可以看成一个双向队列，每次只能从队首或者队尾出队，第n个出队就拿这个数乘以n，最后将和加起来，求最大和

思路：由里向外逆推区间

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[2005],dp[2005][2005];

int main()
{
    int n,i,j,k,l,ans;
    while(~scanf("%d",&n))
    {
        for(i = 1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(i = 1; i<=n; i++)
            dp[i][i] = a[i]*n;//将对角线初始化
        for(l = 1; l<n; l++)
        {
            for(i = 1; i+l<=n; i++)
            {
                j = i+l;
                dp[i][j] = max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]);//这里是从最后出队的开始往前推，之前的初始化也是为了这里，因为只有最后出队的，i+1才会等于j。
            }
        }
        printf("%d
",dp[1][n]);
    }

    return 0;
}