POJ-3186-Treats for the Cows-动态规划DP

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

513152

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.

题解：http://blog.csdn.net/power721/article/details/5803619              http://www.2cto.com/kf/201412/362344.html

区间dp，d[i][j]表示第i个到第j个的最大值

考虑[i, j]这一段，假设其他的都还没拿出来，那么[i, j] 一定由 [i + 1, j] 或者 [i, j - 1] 推得
dp[i][j] = max(dp[i + 1][j] + v[i] * (n - (j - i)), dp[i][j - 1] + v[j] * (n - (j - i)))

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;int dp[2003][2003],a[2003];int main(){int n,i,j;while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        for(i=n;i>0;i--)            for(j=i;j<=n;j++)                dp[i][j]=max(dp[i+1][j]+a[i]*(n+i-j),dp[i][j-1]+a[j]*(n+i-j));        printf("%d\n",dp[1][n]);    }    return 0;}