poj3186（记忆化搜索）

最新推荐文章于 2021-01-27 17:03:08 发布

原创最新推荐文章于 2021-01-27 17:03:08 发布 · 347 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种通过记忆化搜索实现的最优策略，用于解决一个特定问题：如何从序列两端选择元素以获得最大价值。该问题涉及一系列数值，每天可以选择序列头部或尾部的一个数并乘以递增的权重，目标是最大化总价值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Sample Output

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;
}