poj 3186 Treats for the Cows（区间dp）

最新推荐文章于 2021-03-01 20:56:19 发布

cillyb

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： DP 算法 poj

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农场主FJ希望通过最优策略出售一系列储存于双向队列中的奶牛零食，以获取最大收益。零食的价值会随时间增长而提升。本篇介绍了一个动态规划算法实现，通过状态转移方程计算最大可能收入。

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Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5172		Accepted: 2697

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

题意：给出的一系列的数字，可以看成一个双向队列，每次只能从队首或者队尾出队，第n个出队就拿这个数乘以n，最后将和加起来，求最大和

/*
    dp[i][j]记录取i-j能获得的最大价值
    从取最后一个开始想，取最后一个i是a[i]*n, 上次取得就是dp[i-1][i]或dp[i][i+1]
    状态转移方程：dp[i][j] = max(dp[i+1][j]+(n-k)*a[i], dp[i][j-1]+(n-k)*a[j]),  (k是ij区间长度)
*/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 2005;
int a[maxn], dp[maxn][maxn];
int main(void)
{
    int n;
    while(cin >> n)
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]), dp[i][i] = n*a[i];
        for(int k = 1; k < n; k++)
            for(int i = 1; i+k <= n; i++)
            {
                int j = i+k;
                dp[i][j] = max(dp[i+1][j]+(n-k)*a[i], dp[i][j-1]+(n-k)*a[j]);
            }
        printf("%d\n", dp[1][n]);
    }
    return 0;
}