<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>LCA on 111qqz's blog</title><link>https://111qqz.com/en/tags/lca/</link><description>Recent content in LCA on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Thu, 12 Oct 2017 13:44:29 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/lca/index.xml" rel="self" type="application/rss+xml"/><item><title>2016-2017 ACM-ICPC, NEERC, Northern Subregional Contest G Gangsters in Central City (LCA)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-10-12-2016-neerc-subregional-g/</link><pubDate>Thu, 12 Oct 2017 13:44:29 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-10-12-2016-neerc-subregional-g/</guid><description>&lt;p&gt;题意：&lt;/p&gt;
&lt;p&gt;有一棵树，水源在根节点1，房子在叶子节点。有若干操作，操作可能是歹徒占领或者离开一个房子。我们不想给歹徒供水，可以通过切断边实现（如果某个叶子节点到根节点的路径上有一条边被切掉，那么就不能供水了。）对于每次操作后，问不给所有歹徒供水最少要切多少条边，并且问在切满足前面最小的情况下，最少使得多少个良民受影响。初始没有歹徒。&lt;/p&gt;</description></item><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;

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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>leetcode 235. Lowest Common Ancestor of a Binary Search Tree（求一个BST中某两个节点LCA）</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-02-22-leetcode-235-lowest-common-ancestor-of-a-binary-search-tree/</link><pubDate>Wed, 22 Feb 2017 13:22:08 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-02-22-leetcode-235-lowest-common-ancestor-of-a-binary-search-tree/</guid><description>&lt;p&gt;&lt;a href="https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/?tab=Description" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：求一个BST中某两个节点LCA&amp;hellip;.&lt;/p&gt;
&lt;p&gt;思路：卧槽。。。竟然求LCA&amp;hellip;直接想到的显然是Tarjan的方法或者。。。RMQ+DFS。。。但是感觉。。。leetcode怎么可能考算法。。。。于是想到。。。可以从BST下手。。。&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>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 2586 How far away ？ (tarjan算法求LCA模板题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-12-hdu2586/</link><pubDate>Tue, 12 Apr 2016 12:28:42 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-12-hdu2586/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2586" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：一棵树，给出n-1个边权，然后q组查询，每组查询询问两个点之间的距离。
思路：&lt;/p&gt;</description></item></channel></rss>