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Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4259 Accepted: 2150

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.

Source

USACO 2006 February Gold & Silver 

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>

using namespace std;

int m, n, p[10010],dp[2010][2010];

int main()
{
    while (cin >> n)
    {
        for (int i = 1; i <= n; i++)
            cin >> p[i];

        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                dp[i][j] = 0;

        for (int i = 1; i <= n; i++)
            dp[i][i] = p[i] * n;//初始化为最大。

        for (int len = 1; len < n; len++)
        {
            for (int i = 1; i + len <= n; i++)
            {
                int j = i + len; 
                dp[i][j] = max(dp[i + 1][j] + (n - len)*p[i], dp[i][j - 1] + (n - len)*p[j]);//从内向外dp,i j 代表区间范围。
            }
        }
        cout << dp[1][n] << endl;
    }
    return 0;
}