Painting fence 题意 乍一看以为是之前做过的一道单调队列优化的DP,不是. 也是有n块木板,每个木板宽1米,有一个高度ai,现在要把他们刷成橘色,给了你一个宽一米的刷子,你可以横着刷,或者竖着刷,问最少需要刷几下才能将所有的木板着色. 思路 对于一个区间[l,r]的木板来说,第一步要么把所有的木板都竖着刷,要么把最低的木板横着刷一遍,问题变为区间所有的木板减去最短木板的高度之后,刷分为的两个小区间的次数和+最短木板高度.两者取最小值即可.分治 代码 #include<bits/st…

C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his…

memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his favorite color, orange. The fence is represent…

2017-08-02 14:27:18 writer:pprp 题意: • 每块木板宽度均为1,高度为h[i] • n块木板连接为宽度为n的栅栏 • 每次可以刷一横或一竖(上色) • 最少刷多少次可以使得栅栏被全部上色 • 1 ≤ n ≤ 5000 算法分析:可以横着刷,可以竖着刷,横着刷是为了减小竖着刷的次数 采用分治,每个分治中都取横着刷和竖着刷两者的最小值 代码及说明如下: #include <iostream> #include <queue> using namespac…

[题解]Painting Fence 分治模板.贪心加分治.直接\(O(n^2logn)\)分治过去.考虑一块联通的柱形是子问题的,是递归的,贪心分治就可.记得对\(r-l+1\)取\(min\). 上好看的代码 #include<bits/stdc++.h> #define RP(t,a,b) for(register int (t)=(a),edd_=(b);t<=edd_;++t) #define DRP(t,a,b) for(register int (t)=(a),edd_=(…

Codeforces Round #256 (Div. 2) C C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion d…

题目链接:http://codeforces.com/problemset/problem/448/C C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. B…

递归.分治. . . C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his ol…

C. Painting Fence time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output Bizon the Champion isn't just attentive, he also is very hardworking. Bizon the Champion decided to paint his old fence his…

解题报告 给篱笆上色,要求步骤最少,篱笆怎么上色应该懂吧,.,刷子能够在横着和竖着刷,不能跳着刷,,, 假设是竖着刷,应当是篱笆的条数,横着刷的话.就是刷完最短木板的长度,再接着考虑没有刷的木板,,. 递归调用,,. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define inf 999999999999999 using namespace…