poj 3186 Treats for the Cows（区间dp）

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5172		Accepted: 2697

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.

Source

USACO 2006 February Gold & Silver

题意：给出的一系列的数字，可以看成一个双向队列，每次只能从队首或者队尾出队，第n个出队就拿这个数乘以n，最后将和加起来，求最大和

/*
    dp[i][j]记录取i-j能获得的最大价值
    从取最后一个开始想，取最后一个i是a[i]*n, 上次取得就是dp[i-1][i]或dp[i][i+1]
    状态转移方程：dp[i][j] = max(dp[i+1][j]+(n-k)*a[i], dp[i][j-1]+(n-k)*a[j]),  (k是ij区间长度)
*/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 2005;
int a[maxn], dp[maxn][maxn];
int main(void)
{
    int n;
    while(cin >> n)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]), dp[i][i] = n*a[i];
        for(int k = 1; k < n; k++)
            for(int i = 1; i+k <= n; i++)
            {
                int j = i+k;
                dp[i][j] = max(dp[i+1][j]+(n-k)*a[i], dp[i][j-1]+(n-k)*a[j]);
            }
        printf("%d\n", dp[1][n]);
    }
    return 0;
}