POJ 3186 Treats for the Cows （简单区间DP）

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.

题意：

　　 第n次从序列的头部或者尾部取一个数a[i]，求n*a[i]累加和。

题解：

　　很容易想到贪心地前面选取尽量小的数，但是仔细考虑是不行的。前面的选择会对后面的选择产生影响，这时就应该想到DP了。这类问题属于区间DP。假设dp[i][j]是区间[i,j]上的最大值，那么区间dp[i][j]可以由dp[i+1][j]选取a[i]或者dp[i][j-1]选取a[j]得到。但是我们直接dp是不好做的，需要从区间长度入手。具体实现请看代码。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a[2005];
int dp[2005][2005];
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
            cin>>a[i],dp[i][i]=a[i]*n;//区间长度为1的时候，当然是最后一个取。
        for(int len=1;len<n;len++)//枚举区间长度
            for(int i=1;i+len<=n;i++)//枚举区间的起点
            {
                int j=i+len;//区间的终点
                dp[i][j]=max(dp[i+1][j]+(n-len)*a[i],dp[i][j-1]+(n-len)*a[j]);
            }
        cout<<dp[1][n]<<endl;//答案就是1到n这个区间的最大值
    }
    return 0;
}