2022.12.2Treats for the Cows POJ - 3186(区间dp

原题链接：传送门

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

5
1
3
1
5
2

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：

n个数字v(1),v(2),...,v(n-1),v(n)，每次可以取出最左端的数字或者取出最右端的数字，一共取n次取完。如果第i次取的数字是x，可以获得i*x的价值。规划取数顺序，使获得的总价值之和最大 。

思路：

区间dp，定义状态表示为从取到的最大总价值和，每次只能取出最左端的数字或者取出最右端的数字，那么也就是说的状态是由和决定的，所以我们倒着从开始更新到一定是正确的，另外值得注意的是当此时一定是第次取走的状态，所以,由于上一个状态选了也就是说当前是选的第个数。

状态转移方程：

那么AC代码如下：

#include 
int n, a[2010], f[2010][2010];
signed main(){
  scanf("%d", &n);
  for(int i = 1; i <= n; i++){
    scanf("%d", &a[i]);
  }
  for(int i = n; i >= 1; i--){
    for(int j = i; j <= n; j++){
      if(j == i) f[i][j] += n * a[i];
      else f[i][j] += max(f[i + 1][j] + a[i] * (n - j + i), f[i][j - 1] + a[j] * (n - j + i));
    }
  }
  printf("%d\n", f[1][n]);
  return 0;
}