LCA - Lowest Common Ancestor

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A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia 

The lowest common ancestor (LCA) is a concept in graph theory and computer science. Let T be a rooted tree with N nodes. The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself). - Wikipedia

Your task in this problem is to find the LCA of any two given nodes v and w in a given tree T.

For example the LCA of nodes 9 and 12 in this tree is the node number 3.


The first line of input will be the number of test cases. Each test case will start with a number N the number of nodes in the tree, 1 <= N <= 1,000. Nodes are numbered from 1 to N. The next N lines each one will start with a number M the number of child nodes of the Nth node, 0 <= M <= 999 followed by M numbers the child nodes of the Nth node. The next line will be a number Q the number of queries you have to answer for the given tree T, 1 <= Q <= 1000. The next Q lines each one will have two number v and w in which you have to find the LCA of v and w in T, 1 <= v, w <= 1,000.

Input will guarantee that there is only one root and no cycles.


For each test case print Q + 1 lines, The first line will have “Case C:” without quotes where C is the case number starting with 1. The next Q lines should be the LCA of the given v and w respectively.


3 2 3 4
3 5 6 7
5 7
2 7

Case 1:

hide comments
mrinal_verma: 2017-07-25 21:41:25

can someone plz tell me what is wrong in my code , im using lca+rmq

Last edit: 2017-07-26 09:17:26
sonu: 2017-06-15 11:11:29

build ->o(n)
query->o(sqrt(n)) works!!

rraj001: 2017-05-30 11:03:13

My 100th :)

anupamwadhwa: 2017-05-15 01:10:57

TAKE "1" as ROOT node always.

I take root node as node whose parent is not given in input.--> WA

Andres R. Arrieche S. [UCLA-ve]: 2017-04-25 17:37:09

Same algorithm:
C++ AC with time 0.37 .
Java TLE

ashu121: 2017-03-25 00:48:08

AC in one go!
just make level of both nodes same and then trace their common parent :)

cake_is_a_lie: 2017-03-02 06:05:38

I just used a single set of additional back-pointers (BIT-like) and a level tag, 0.02s. Much faster to code than <O(N), O(1)> solution.

darshan_7807: 2016-12-06 12:02:42

LCA with RMQ

hamjosh1: 2016-12-04 00:02:37

Node 1 is always the root :D

xin đừng quên tôi: 2016-09-28 16:48:36


Added by:hossamyosef
Time limit:0.600s-1.113s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Resource:FCIS/ASU Local Contest 2013