POJ 3186 Treats for the Cows

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4748		Accepted: 2447

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.

题意是给定一串数字，每次只能从头或者尾输出，每次输出都有一个权值，且权值从1到N递增。问怎么取得到的最后的值最大。

首先要确定dp数组，我们可以设置dp[i][j]表示从i为第一个数字到j为最后一个数字的这一串数字对应的最大的值。

很容易就可以得出 dp[i][j] = max(dp[i + 1][j] +  a[i] * (N - j  + i), dp[i][j - 1] + a[j] * (N - j + i))

也就是比较下一次从两边取那个更大。

代码如下：

/*************************************************************************
	> File Name: Treats_for_the_Cows.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Wed 28 Oct 2015 03:11:28 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
int N;
int dp[2010][2010];
int a[2010];
int main(void){
    scanf("%d", &N);
    for(int i  = 1; i <= N; i ++){
        scanf("%d", &a[i]);
        dp[i][i] = a[i];
    }

    for(int i = N; i > 0; i --){
        for(int j  = i; j <= N; j ++){
            dp[i][j] = max(dp[i + 1][j] + a[i] * (N - j + i), dp[i][j - 1] + a[j] * (N - j + i));
        }
    }

    printf("%d
", dp[1][N]);
    return 0;
}