POJ3186 DP

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5753		Accepted: 2972

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：

n个数排成一串，每次可以从头取一个数或者从尾取一个数，取出数后该数的值乘以取他的时间（每取一个数时间加一）是价值，问能够得到的最大价值。

代码：

//dp，要从内向外推，dp[i][j]表示串中只剩下i~j数的时候最大价值
//dp[i][j]=max(dp[i+1][j]+a[i]*cnt,dp[i][j-1]+a[j]*cnt),因为dp[i][j]时要用到
//后面的数，要从近距离到远距离dp。当然也可以从前向后。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,a[];
int f[][];
int main()
{
    while(scanf("%d",&n)==){
        for(int i=;i<n;i++) scanf("%d",&a[i]);
        for(int i=n-;i>=;i--){
            for(int j=i;j<n;j++){
                f[i][j]=max(f[i+][j]+a[i]*(n+i-j),f[i][j-]+a[j]*(n+i-j));
            }
        }
        printf("%d\n",f[][n-]);
    }
    return ;
}