SP14932 LCA - Lowest Common Ancestor-LCA

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本文深入解析了在给定树结构中寻找两个节点的最近公共祖先（LCA）的问题，通过具体示例和算法实现，详细介绍了如何构建和遍历树形结构以高效解决LCA问题。

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SP14932 LCA - Lowest Common Ancestor

Description:

一棵树是一个简单无向图，图中任意两个节点仅被一条边连接，所有连通无环无向图都是一棵树。-Wikipedia

最近公共祖先（LCA）是……（此处省去对LCA的描述），你的任务是对一棵给定的树TT以及上面的两个节点u,vu,v求出他们的LCALCA。

例如图中99和1212号节点的LCALCA为33号节点

Input:

输入的第一行为数据组数TT，对于每组数据，第一行为一个整数N(1\leq N\leq1000)N(1≤N≤1000)，节点编号从11到NN，接下来的NN行里每一行开头有一个数字M(0\leq M\leq999)M(0≤M≤999)，MM为第ii个节点的子节点数量，接下来有MM个数表示第ii个节点的子节点编号。下面一行会有一个整数Q(1\leq Q\leq1000)Q(1≤Q≤1000)，接下来的QQ行每行有两个数u,vu,v，输出节点u,vu,v在给定树中的LCALCA。

输入数据保证只有一个根节点并且没有环。

Output:

对于每一组数据输出Q+1Q+1行，第一行格式为"Case i:"（没有双引号），i表示当前数据是第几组，接下来的QQ行每一行一个整数表示一对节点u,vu,v的LCALCA。

Sample Input:

Sample Output:

Case 1:
3
1

Translated by @yxl_gl

题目描述

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.

Input

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.

Output

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.

Example

Input:
1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7

Output:
Case 1:
3
1

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+6;
bool vis[maxn];
int pre[maxn];
struct node{
	int to;
	int from;
	int LCA;
	int next;
}sz[maxn],sr[maxn];
int numadd1,numadd2;
int q=1;
int head[maxn],ahead[maxn];
int find(int x)
{
	return x==pre[x]?x:pre[x]=find(pre[x]);
}
void add(int from ,int to)
{
	sz[++numadd1].next=head[from];
	sz[numadd1].to=to;
	head[from]=numadd1;
}
void add1(int from,int to)
{
	sr[++numadd2].next=ahead[from];
	sr[numadd2].to=to;
	ahead[from]=numadd2;
}
void dfs(int x)
{
//	cout<<x<<endl;
	pre[x]=x;
	vis[x]=1;
	for(int i=head[x];i;i=sz[i].next)
	{
		if(vis[sz[i].to]==0)
		{
			dfs(sz[i].to);
			pre[sz[i].to]=x;
		}
	}
	for(int i=ahead[x];i;i=sr[i].next)
	{
		if(vis[sr[i].to])
		{
			sr[i].LCA=find(sr[i].to);
			if(i&1)	sr[i+1].LCA=sr[i].LCA;
			else sr[i-1].LCA=sr[i].LCA;
		}
	}
}
void init()
{
	memset(vis,0,sizeof vis);
//	memset(sz,0,sizeof sz);
//	memset(sr,0,sizeof sr);
	memset(head,0,sizeof head);
	memset(ahead,0,sizeof ahead);
	numadd1=0,numadd2=0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{	
		init();//初始化 
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)//遍历给出边的联系 
		{
			int x;
			scanf("%d",&x);
			if(x==0)	continue;
			else
			{
				for(int j=0;j<x;j++)
				{
					int y;
					scanf("%d",&y);
					add(i,y);
					add(y,i);	
				}	
			} 
		}
		int m;
		scanf("%d",&m);
		for(int i=0;i<m;i++)
		{
			int x,y;
			scanf("%d %d",&x,&y);
			add1(x,y);
			add1(y,x);
		}
		dfs(1);
		printf("Case %d:\n",q++);
		for(int i=1;i<=m;i++)
		{
			printf("%d\n",sr[i*2].LCA);
		}
	}
}