POJ3186：Treats for the Cows(区间DP)

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

思路：先求距离小的

#include <cstdio>#include <cstdlib>#include <string.h>#include <string>#include <math.h>#include <algorithm>#include <iostream>#include <queue>#include <stack>#include <map>#include <set>using namespace std;int i,j,d,t;int a[2010], s[2010], dp[2010][2010];int main(){    cin>>t;    for(i=0;i<t;i++)    cin>>a[i];    for(i=1;i<=t;i++)    s[i]=s[i-1]+a[i-1];    for(i=0;i<t;i++)    dp[i][i]=a[i];    for(d=1;d<t;d++)    for(i=0;i+d<t;i++)    {    dp[i][i+d]=max(dp[i+1][i+d]+s[i+d+1]-s[i+1]+a[i], dp[i][i+d-1]+s[i+d]-s[i]+a[i+d]);    }    cout << dp[0][t-1] <<endl;    return 0;}