POJ_3186_Treats for the Cows

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6568		Accepted: 3459

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

 

• 最开始想的是贪心
• 每次取两端最小的，这样把大的留在最后，总和最大
• 但是WA
• 原因在于贪心的局限性上，我们不能保证这样的贪心策略在应对诸如当前右端比左端大但是右端第二个数比两端点都小的情况下怎样取舍的情形下怎样做
• 所以还是应该dp
• 不难看出每一个点在双端队列中的出队顺序排除开始的两端是1到n，其他的都是2到n，所以我们在确定一定区间的最优值之后可以向两边递推

 #include <iostream>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <climits>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long           LL ;
 typedef unsigned long long ULL ;
 const int    maxn = 2e3 +   ;
 const int    inf  = 0x3f3f3f3f ;
 const int    npos = -         ;
 const int    mod  = 1e9 +     ;
 const int    mxx  =  +     ;
 const double eps  = 1e-       ;
 const double PI   = acos(-1.0) ;

 int dp[maxn][maxn], a[maxn], n;
 int main(){
     // freopen("in.txt","r",stdin);
     // freopen("out.txt","w",stdout);
     while(~scanf("%d",&n)){
         for(int i=;i<=n;i++){
             scanf("%d",&a[i]);
             dp[i][]=a[i]*n;
         }
         for(int j=;j<n;j++)
             for(int i=n-j;i>;i--)
                 dp[i][j]=max(dp[i][j-]+a[i+j]*(n-j),dp[i+][j-]+a[i]*(n-j));
         printf("%d\n",dp[][n-]);
     }
     return ;
 }