POJ-3186-DP

Treats for the Cows

Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

SubmitStatus

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

513152

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.
刚开始是用数组模拟队列写的，用两个变量i,j标记左面和右面的出列的位置，就是比较以i,j为下标的数组的大小，如果i<=j，那么就a[i]出列，并且i++，但是WA
想不明白为什么要用DP，后来才想到我那样做只能保证眼前能达到收益最大，并不能保证总的收益最大，比如1,7,1,1,3,2这个例子
DP代码如下：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[2005],dp[2005][2005];
int main()
{
    int n,i,l,j,k,ans;
    while(~scanf("%d",&n))
    {
        for(i=1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
            dp[i][i]=a[i]*n;
        for(l=1; l<n; l++)
        {
            for(i=1; i+l<=n; i++)
            {
                j=i+l;
                dp[i][j]=max(dp[i+1][j]+(n-l)*a[i],dp[i][j-1]+(n-l)*a[j]);
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}
这个是照网上打的，但是发现总有错，原来是1和l有的地方没有分清楚