斜率优化dp简讲 && HDU 3507 Print Article

最新推荐文章于 2020-05-21 21:25:44 发布

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最新推荐文章于 2020-05-21 21:25:44 发布 · 3.1k 阅读

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#dp #斜率优化 #优化

Problem Description

Zero has an old printer that doesn’t work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

这里写图片描述

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

Output

A single number, meaning the mininum cost to print the article.

题解：
其实想学斜率优化很久了，本来想等省选后学的，果然省选就考到了。真是有趣。

HDU 3507 算是斜率优化的经典题目了，在网上看到很多大神都用这道题作为例题，零零散散看了两天才把这道题A了。下面先来分析一下这道题再简要说一下斜率优化。

首先这道题的dp方程不难看出：
f[i]=min( f[j]+( sum[i]-sum[j] )^2 )+M
f[i]表示如果从第i个数后面断开的最优解，假设第j个数是上一个断点，那么此时的解就是f[j]+( sum[i]-sum[j] )^2 +M；所以我们枚举前i-1个断点，取最优解。这样做简直跟暴力没什么区别，暴力枚举，，n^2的复杂度，500000的数据很显然是不可以的。那么我们接下来分析一下我们的dp方程：
f[i]=min(