POJ3186Treats for the Cows题解动态规划DP

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2686 Accepted: 1280

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.

Source

USACO 2006 February Gold & Silver

状态：

d[i][j]表示第i个到第j个的最大值

状态转移方程：

d[i][j]=max(d[i+1][j]+a[i]*(n+i-j),d[i][j-1]+a[j]*(n+i-j))

为了满足无后效性i逆序循环

代码：

#include<stdio.h>#define max(a,b) ((a)>(b)?(a):(b))int dp[2002][2002];int main(){int n,i,j,a[2002];scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",a+i);for(i=n;i>0;i--)//关键：倒着循环for(j=i;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));printf("%d/n",dp[1][n]);}