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.

/*
题意：给你一个双向队列，每次可以从队首，或者从队尾取出元素，每次操作会获得相应的价值，第i个取出的元素a
    得到的价值就是i*a，问你能取出的最大价值是多少

初步思路：先贪心搞一发试试

#改进：用一个数组b 存放着 逆序的数组 a，dp[i][j]表示，a数组取i 个，b数组取 j个时的最大值，得到状态转移方程：
    dp[i][j]=max(dp[i-1][j]+a[i]*(i+j),dp[i][j-1]+b[j]*(i+j)); 实际上就是 左边取 i 个，右边取 j
    个的最大值

#错误：上面的初始化，没法搞，当i等于零或者j等于零的时候，问题就重新转化为题目要求的问题了，换一个思路，从里面向
    外边扩，不从边上开始，从里往外，因为最终的顶点并不是确定的，dp[i][j]表示从i到j能得到的最大价值，得到状态
    转移方程： dp[i][j]=max( dp[i+1][j]+a[i]*( n-j+i ), dp[i][j-1]+a[j]*( n-j+i ) ); 

#错误：得不到正确的结果

#注意：上面的想法没错，但是i是逆向循环的，因为正向循环的话，记录的状态都是接下来的状态，根本没有参考上一个状态，
    进行判断结果
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int n;
int l,r;
int a[];
int dp[][];
void init(){
    memset(dp,,sizeof dp);
    memset(a,,sizeof a);
}
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
            dp[i][i]=a[i]*n;//初始化每一个最后拿的话都是，a[i]*n
        }
        for(int i=n-;i>=;i--){
            for(int j=i+;j<=n;j++){
                dp[i][j]=max( dp[i+][j]+a[i]*( n-j+i ), dp[i][j-]+a[j]*( n-j+i ) );
                // cout<<"( "<<i<<" ,"<<j<<" ) "<<( n-j+i )<<endl;
            }
        }
        // for(int i=1;i<=n;i++){
        //     for(int j=1;j<=n;j++){
        //         cout<<dp[i][j]<<" ";
        //     }
        //     cout<<endl;
        // }
        printf("%d\n",dp[][n]);
    }
    return ;
}