POJ-3186_Treats for the Cows

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536K

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.

题意：给予一个数组，每次可以取前面的或者后面的，第k次取的v[i]价值为v[i]*k，问总价值最大是多少。
题解：一个区间DP题目，每一次取的时候可以由d[i+1][j]或者d[i][j-1]转移而来。
转移方程：dp[i][j]=max(dp[i+1][j]+p[i]*(n+i-j),dp[i][j-1]+p[j]*(n+i-j)); 其中n-(j-i)是第几次取。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>

using namespace std;

const int maxn = 2050;
int dp[maxn][maxn];

int main()
{
    int n,v[maxn],i,j;
    memset(dp,0,sizeof(dp));
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        {
            scanf("%d",&p[i]);
            dp[i][i]=p[i];
        }
    for(i=n;i>=1;i--)
        for(j=i;j<=n;j++)
            dp[i][j] = max(dp[i+1][j]+v[i]*(n-(j-i)),dp[i][j-1]+v[j]*(n-(j-i)));
    printf("%d\n",dp[1][n]);
    return 0;
}