poj3186（记忆化搜索）

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. 

题目大意就是给一个序列，每次从序列头或序列尾取出一个数字乘以当前的权重，求能取到的最大值是多少

直接记忆化搜索，n平方的复杂度，太舒服了

AC代码

#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<cstring>using namespace std;int num[2005];int dp[2005][2005]={0};int dfs(int sta,int End,int now){    if(dp[sta][End]&&sta!=End)return dp[sta][End];    if(dp[sta][End])return dp[sta][End]*now;//取出来的不是区间而是单个数字的时候才乘以权重    int ans=max(dfs(sta,End-1,now+1)+dp[End][End]*now,dfs(sta+1,End,now+1)+dp[sta][sta]*now);    return dp[sta][End]=ans;}int main(){    int N;    while(~scanf("%d",&N))    {        for(int i=1;i<=N;i++)scanf("%d",&dp[i][i]);        printf("%d\n",dfs(1,N,1));    }    return 0;}