POJ 3186Treats for the Cows 递推

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5234 Accepted: 2737

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

很简单的题，一步递推就出来了...

#include <iostream>#include <cstring>#include <algorithm>using namespace std;int dp[2005][2005];int main(){    int n,a[2005];    while(cin>>n)    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            cin>>a[i];        dp[1][1]=a[1];        dp[1][0]=a[n];        for(int i=2;i<=n;i++)        {            for(int j=0;j<=i;j++)            {                if(j!=i)                    dp[i][j]=max(dp[i][j],dp[i-1][j]+i*a[n-i+j+1]);                if(j!=0)                    dp[i][j]=max(dp[i][j],dp[i-1][j-1]+i*a[j]);            }        }        cout<<*max_element(dp[n],dp[n]+n+1)<<endl;    }    return 0;}