POJ3186 Treats for the Cows DP

Treats for the Cows

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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
문제풀이: 제시된 일련의 숫자는 양방향 대열로 볼 수 있다. 매번 대열의 첫 번째 또는 대열의 끝에서만 대열을 나갈 수 있다. n번째 대열은 이 수를 n에 곱하고 마지막에 합쳐서 최대와

코드

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

int dp[2005][2005];
int a[2005];
int main()
{
   int n;
   while(~scanf("%d",&n)){
   memset(dp,0,sizeof(dp));
   for (int i = 1;i<=n;i++)
   {
       scanf("%d",&a[i]);
   }
   for (int i = 1;i<=n;i++)
      dp[i][i] = a[i]*n;
   int j;
   for (int k = 1;k<n;k++)
   {
     for (int i = 1;k+i<=n;i++)
     {
          j = i+k;
          dp[i][j] = max(dp[i+1][j]+(n-k)*a[i],dp[i][j-1]+(n-k)*a[j]);
     }
   }
   printf("%d
",dp[1][n]);
   }
}