Treats for the Cows

Treats for the Cows

 Treats for the Cows

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3186

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.

本题是一个区间DP问题，如果我们按照常规思路的话，会无法确定最终状态，所以，我们可以逆向思维一下，让区间由内向外扩。首先我们每个元素都有可能是最后一次拿的，所以我们让dp[i][i]=v[i]*n。然后我们再遍历每一个可能的区间，每次取最优解，最后当区间扩到dp[1][n]的时候，我们认为结束，此时的dp[1][n]即为我们所求结果。状态转移方程为dp[i]=max{dp[i+1][j]+v[i]*(n-j+i),dp[i][j-1]+v[j]*(n-j+i)}.注意 ：n-j+i为天数；i要逆序，j顺序。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[2005][2005];
int main()
{
    int n,v[2005];
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&v[i]);
        dp[i][i]=v[i]*n;
    }
    for(int i=n-1;i>=1;i--)
    {
        for(int j=i+1;j<=n;j++)
        {
            dp[i][j]=max(dp[i+1][j]+v[i]*(n-j+i),dp[i][j-1]+v[j]*(n-j+i));
        }
    }
    printf("%d\n",dp[1][n]);
    return 0;
}