poj 3186 Treats for the Cows(区间dp)

最新推荐文章于 2020-03-05 23:55:10 发布

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最新推荐文章于 2020-03-05 23:55:10 发布

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分类专栏： POJ dp 区间dp

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这篇博客探讨了如何运用区间动态规划（dp）策略解决‘Treats for the Cows’问题。题目中提到，FJ的零食按编号存放在一端开口的盒子里，每天可以从任一端取出一个零食。零食的价值随时间增加，且价值不等。目标是找出在特定时间内，为了获取最大价值，FJ应如何选择取出零食的顺序。输入和输出格式以及样例输入输出也进行了说明。

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

solution：

dp[i][j]代表从i~j需要的最大花费，我们可以推出计算i~j时，时间为n-j+i天,因此递推公式为dp[i][j]=max(dp[i+1][j]+v[i]*(n-j+i),dp[i][j-1]+v[j]*(n-j+i))

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int dp[2200][2200],v[2200];
int n;
int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++)
			scanf("%d", &v[i]);
		memset(dp, 0, sizeof(dp));
		for (int i = n; i >= 1; i--)
			for (int j = i; j <= n; j++)
				dp[i][j] = max(dp[i + 1][j] + v[i] * (n - j+i), dp[i][j - 1] + v[j] * (n-j+i));
		printf("%d\n", dp[1][n]);
	}
}