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\n", dp[1]
);
return 0;
}