Poj 3186 Treats for the Cows【区间dp】

最新推荐文章于 2021-03-01 20:56:19 发布

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最新推荐文章于 2021-03-01 20:56:19 发布

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分类专栏： dp 文章标签： Poj 3186

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农场主FJ拥有N个美味的奶牛奖励品，每天从两端取出一个进行售卖以获得最大收益。本篇介绍如何通过动态规划算法找到最佳销售顺序。

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Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5427		Accepted: 2825

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

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.

Source

USACO 2006 February Gold & Silver

题目大意：

给你N个物品，我们每次只能从两边拿，对于第i个拿出来的物品，那么其价值就增加i*这个物品的价值。

问拿取的最大价值是多少。

思路（这个区间dp好水啊）：

1、设定dp【i】【j】表示从i到j的拿取的最大的价值是多少。

2、我们考虑逆向考虑，那么我们对于区间【i，j】，可以是区间【i，j-1】再添加一个a【j】得来，也可以是区间【i+1，j】添加一个a【i】得来。

那么对应有：

①dp【i】【j】=max（dp【i】【j】，dp【i】【j-1】+（i+n-j）*a【j】）；

②dp【i】【j】=max（dp【i】【j】，dp【i+1】【j】+（i+n-j）*a【i】）；

3、ans=dp【1】【n】；

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
ll dp[2005][2005];
ll a[2005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        memset(dp,0,sizeof(dp));
        for(int len=0;len<=n;len++)
        {
            for(int i=1;i+len<=n;i++)
            {                
                int j=i+len;
                if(len==0)
                {
                    dp[i][j]=a[i]*n;
                }
                else
                {
                    dp[i][j]=max(dp[i][j],dp[i][j-1]+(i+n-j)*a[j]);
                    dp[i][j]=max(dp[i][j],dp[i+1][j]+(i+n-j)*a[i]);
                }
            }
        }
        printf("%I64d\n",dp[1][n]);
    }
}