POJ 3186 Treats for the Cows （区间DP）

Treats for the Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5094   Accepted: 2654 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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn= 2000;
const int INF = 0x3f3f3f3f;
int t,n;
int a[maxn];
int dp[maxn][maxn];

int main()
{
cin>>n;
memset(dp,0,sizeof dp);
for(int i=1;i<=n;i++){
cin>>a[i];
dp[i][i]=n*a[i];
}
for(int i=n;i>=1;i--){
for(int j=i+1;j<=n;j++){
dp[i][j]=max(dp[i+1][j]+a[i]*(n+i-j),dp[i][j-1]+a[j]*(n+i-j));
}
}
cout<<dp[1]
<<endl;
}