<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Rmq on 111qqz's blog</title><link>https://111qqz.com/en/tags/rmq/</link><description>Recent content in Rmq on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Sun, 30 Jul 2017 17:45:18 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/rmq/index.xml" rel="self" type="application/rss+xml"/><item><title>hdu 3078 Network (LCA)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-07-30-hdu-3078/</link><pubDate>Sun, 30 Jul 2017 17:45:18 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-07-30-hdu-3078/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=3078" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;

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&lt;p&gt;一棵树，给出点权，问一条树链上第k大的点权，点权可以动态修改。&lt;/p&gt;</description></item><item><title>codeforces #425 D. Misha, Grisha and Underground (dfs+rmq在线求LCA,讨论了一年)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-07-30-codeforces-div2-425d/</link><pubDate>Sun, 30 Jul 2017 12:05:30 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-07-30-codeforces-div2-425d/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/contest/832/problem/D" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;

&lt;h2 class="relative group"&gt;题意：
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&lt;p&gt;给出一棵树，以及三个点（可能重合），问两两组成的3条路径中，哪2条路径重合部分最长。&lt;/p&gt;</description></item><item><title>codeforces 123 D. String　（后缀数组+两次二分得到区间＋rmq）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-08-02-cf-123/</link><pubDate>Tue, 02 Aug 2016 15:23:55 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-08-02-cf-123/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/problemset/problem/123/D" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：定义一个函数F．．&lt;/p&gt;
&lt;blockquote&gt;&lt;p&gt;For exampe: &lt;em&gt;F&lt;/em&gt;(&lt;em&gt;babbabbababbab&lt;/em&gt;, &lt;em&gt;babb&lt;/em&gt;) = 6. The list of pairs is as follows:&lt;/p&gt;
&lt;p&gt;(1, 4), (4, 7), (9, 12)&lt;/p&gt;</description></item><item><title>hdu 4123 Bob’s Race (树的直径+尺取+rmq)(珍爱生命，远离log)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-17-hdu-4123/</link><pubDate>Sun, 17 Jul 2016 16:21:03 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-17-hdu-4123/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4123" target="_blank" rel="noreferrer"&gt;hdu 4123 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：一棵树，定义d[i]为点i到树上某点的最大距离。。。给出若干查询，每个查询一个x,问最多能有多少点满足这些点中，最大的d与最小的d的差小于等于x.要求这些点的编号必须是连续的。&lt;/p&gt;</description></item><item><title>zoj 3195 Design the city （lca,dfs+rmq）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-21-zoj-3195/</link><pubDate>Sat, 21 May 2016 08:31:20 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-21-zoj-3195/</guid><description>&lt;p&gt;&lt;a href="http://www.icpc.moe/onlinejudge/showProblem.do?problemId=3320" target="_blank" rel="noreferrer"&gt;zoj 3195题目链接&lt;/a&gt;
题意：求树上三点的最短距离。。。
思路：两两求，和除以2.
因为忘记初始化p=0..WA了将近两个小时。。。？
妈的智障。&lt;/p&gt;</description></item><item><title>hdu 2874 Connections between cities (添加虚点，并查集+LCA(rmq+dfs))</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-21-hdu-2874/</link><pubDate>Sat, 21 May 2016 04:58:22 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-21-hdu-2874/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2874" target="_blank" rel="noreferrer"&gt;hdu2874题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给一个森林，问两点的最短距离，或者输出两点不联通。&lt;/p&gt;
&lt;p&gt;思路：&lt;strong&gt;最最重要的一点是:添加虚点！&lt;/strong&gt;&lt;/p&gt;</description></item><item><title>poj 1986 Distance Queries (lca,在线做法dfs+rmq)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-20-poj-1986/</link><pubDate>Fri, 20 May 2016 11:25:49 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-20-poj-1986/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=1986" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：求树上两点的最短距离？
思路： dis[i]表示点i到根节点的距离，那么任意两点u,v的最短距离d = dis[u]+dis[v]-2*dis[LCA(u,v)].
只需要求出rmq+dfs的在线方法求出lca(u,v)即可。&lt;/p&gt;</description></item><item><title>hdu 3530 Subsequence (尺取+rmq)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-hdu3530/</link><pubDate>Thu, 19 May 2016 09:44:36 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-hdu3530/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=3530" target="_blank" rel="noreferrer"&gt;hdu 3530题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出n个数，m,k，问最大的j-i+1,使得【i,j】间的最大值与最小值的差属于[m,k]
思路：rmq+尺取。 2A.&lt;/p&gt;</description></item><item><title>poj 1470 Closest Common Ancestors (lca,rmq+dfs,读入技巧)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-poj1470/</link><pubDate>Thu, 19 May 2016 08:42:05 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-poj1470/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=1470" target="_blank" rel="noreferrer"&gt;poj1470题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：求两点的lca.
思路：dfs+rmq. 读入技巧。
读入比较坑爹。。。
学会了一种新的读入技巧。&lt;/p&gt;</description></item><item><title>poj 1330 Nearest Common Ancestors (lca,用dfs+rmq在线求解)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-poj1330/</link><pubDate>Thu, 19 May 2016 07:39:48 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-19-poj1330/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=1330" target="_blank" rel="noreferrer"&gt;poj1330题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出一棵树，求两点的lca.
思路：将lca转化成rmq在线求解。&lt;/p&gt;</description></item><item><title>hdu 4122 Alice's mooncake shop(rmq)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-hdu4122/</link><pubDate>Wed, 18 May 2016 13:03:34 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-hdu4122/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4122" target="_blank" rel="noreferrer"&gt;hdu2142题目链接&lt;/a&gt;
题意：有n个订单和可以在m小时内制作月饼
接下来是n个订单的信息:需要在mon月,d日,year年,h小时交付订单r个月饼
接下来一行t,s表示制作的月饼可以保质t天，每保质一天需要花费s的价值
接下来m行表示从第0小时开始在该时间制作月饼的花费的价值
求完成所有订单消耗的最小价值&lt;/p&gt;</description></item><item><title>poj 3368 Frequent values （暴力+rmq，分类讨论）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-poj3368/</link><pubDate>Wed, 18 May 2016 07:47:48 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-poj3368/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=3368" target="_blank" rel="noreferrer"&gt;poj 3368 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出n个非减的数a[i],求区间[l,r]中出现次数最多的数的出现的次数。&lt;/p&gt;</description></item><item><title>poj 2452 Sticks Problem (rmq+二分，需要返回最值位置)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-poj2452/</link><pubDate>Wed, 18 May 2016 05:40:40 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-18-poj2452/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=2452" target="_blank" rel="noreferrer"&gt;poj2452题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给你一组数a[n]，求满足a[i] &amp;lt; a[k] &amp;lt; a[j] (i &amp;lt;= k &amp;lt;= j)的最大的j-i。&lt;/p&gt;</description></item><item><title>lightoj 1081 Square Queries (二维rmq，降维)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-loj1081/</link><pubDate>Mon, 16 May 2016 12:06:33 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-loj1081/</guid><description>&lt;p&gt;&lt;a href="http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26798" target="_blank" rel="noreferrer"&gt;lightoj 1081 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：和上一道一样，但是由于size变成了500，如果按照之前的做法会tle + mle&amp;hellip;&lt;/p&gt;</description></item><item><title>poj 2019 Cornfields (二维rmq)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-poj2019/</link><pubDate>Mon, 16 May 2016 11:01:34 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-poj2019/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=2019" target="_blank" rel="noreferrer"&gt;poj2019题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给一个方阵，k个查询，每个查询求某个方阵的最大值和最小值之差。&lt;/p&gt;</description></item><item><title>hdu 2888 check corners (二维rmq模板题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-hdu2888/</link><pubDate>Mon, 16 May 2016 10:43:05 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-16-hdu2888/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2888" target="_blank" rel="noreferrer"&gt;hdu2888题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：问某个矩阵内的最大值，并且问最大值是否是在四个角中出现。
思路：二维rmq.需要注意数组稍微开大1就会MLE,因为是四维数组，一维大一点，整个就会大很多==。&lt;/p&gt;</description></item><item><title>BZOJ 1636: [Usaco2007 Jan]Balanced Lineup (RMQ模板题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-05-15-bzoj1636/</link><pubDate>Sun, 15 May 2016 13:31:30 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-05-15-bzoj1636/</guid><description>&lt;h2 class="relative group"&gt;1636: [Usaco2007 Jan]Balanced Lineup
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&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 680  Solved: 493
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1636" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1636" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1636" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item></channel></rss>