poj3186Treats 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.

Line 1: A single integer, N 

Lines 2..N+1: Line i+1 contains the value of treat v(i)

Line 1: The maximum revenue FJ can achieve by selling the treats

5
1
3
1
5
2

43

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.

只能两端取出。

#include<stdio.h>
int dp[2005][2005];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int n,a[2005],ans[2005];
    while(scanf("%d",&n)>0)
    {
        ans[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            ans[i]=ans[i-1]+a[i];
            for(int j=1;j<=n;j++)
            dp[i][j]=0;
        }
        for(int r=0;r<n;r++)
        for(int i=1;i<=n-r;i++)
        {
            int j=i+r;
            dp[i][j]=max(dp[i][j-1]+a[j]+ans[j-1]-ans[i-1],a[i]+ans[j]-ans[i]+dp[i+1][j]);
        }
        printf("%d\n",dp[1][n]);
    }
}