POJ - 3186 Treats for the Cows （区间dp）

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：

有 n 个数排成一行，每次可以取左右两端的数，第 k 次取数的贡献是 k * a[i]，求最大贡献值

思路：

这里用逆过程，即从这一行数中取出一个数，剩下的数必须在取出的数的左右两侧取。dp[i][j]表示区间 [i, j] 都取完得到的最大值，可以由dp[i + 1][j]和dp[i][j - 1]推过来

#include <iostream>
#include <cstdio>
#include <map>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const int N = 2e3 + 10;

int a[N], dp[N][N];

int main()
{
    int n, t;
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d", &a[i]);
            dp[i][i] = a[i] * n;    //先把每个都当成最后一个取走的，贡献值为 n 倍
        }
        for(int i = n; i >= 1; --i)//枚举起点
        {
            for(int j = i + 1; j <= n; ++j)//枚举终点
            {
                dp[i][j] = max(dp[i + 1][j] + a[i] * (n - j + i), dp[i][j - 1] + a[j] * (n - j + i));    //一种是最近取的是a[i]，一种是最近取的a[j]
            }
        }
        cout<<dp[1][n]<<'\n';    //整个区间都取完了
    }
    return 0;
}