2022.12.2Treats for the Cows POJ

2022.12.2Treats for the Cows POJ - 3186(区间dp

news2024/11/15 4:39:43

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

Inputcopy	Outputcopy
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 $dp[i][j]$ 状态表示为从 $i$ 取到 $j$ 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 $dp[i][j]$ 的状态是由 $dp[i+1][j]$ 和 $dp[i][j-1]$ 决定的，所以我们倒着从 $dp[n][n]$ 开始更新到 $dp[1][n]$ 一定是正确的，另外值得注意的是当 $i==j$ 此时一定是第 $n$ 次取走的状态，所以 $dp[i][i]=n\times a[i]$ ,由于上一个状态选了 $j-i$ 也就是说当前是选的第 $n-j+i$ 个数。

状态转移方程：

$dp[i][j]=\left\{\begin{matrix} n\times a[i] ,i==j& \\dp[i+1][j]+(n-j+i)\times a[i]+dp[i][j-1]+(n-j+i)\times a[j],i\neq j& \end{matrix}\right.$

那么AC代码如下：

#include <iostream>
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;
}