[倍增 LCA]

题目描述 
给定n个点的树(1是根)，m次询问，每次询问两点的LCA;

输入格式 
第一行两个整数,n,m;

接下来n-1行,给定正整数a,b,表示a,b间有边;

接下来m行,给定整数a,b,表示询问a,b;

输出格式 
对于每个询问,输出它们的LCA;

样例数据 
input

3 2 
1 2 
2 3 
1 2 
2 3 
output

1 
2 
数据规模与约定 
保证所有数据n=m=100000

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100010;

int n, m;
int to[maxn], fr[maxn], edge[maxn];
int fa[maxn][21];
int dep[maxn];

void dfs(int id, int deep, int f)
{
	fa[id][0] = f;
	dep[id] = deep;
	for (int i = edge[id]; i != -1; i = fr[i])
	{
		int To = to[i];
		//cout<<To <<' '<< deep + 1<<' '<<id<<endl;
		dfs(To, deep + 1, id);
	}
	return;
}

void beizeng()
{
	for (int j = 1; j <= 20; j++)
	{
		for (int i = 0; i <= n; i++)
		{
			int t = fa[i][j - 1];

			if (t == -1)
				fa[i][j] = -1;
			else
				fa[i][j] = fa[t][j - 1];
		}
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	memset(edge, -1, sizeof edge);
	for (int i = 1; i <= n - 1; i++)
	{
		int a, b;
		scanf("%d%d", &a, &b);

		to[i] = b;
		fr[i] = edge[a];
		edge[a] = i;
	}

	dfs(1, 0, -1);
	for (int i = 1;i <= n; i++)
	{
		for (int j = 0; j <= 20; j++)
			cout<<fa[i][j]<<' ';
		cout<<endl;
	}
	beizeng();   // fa[i][j] 打表 O(n * log[n])
	
	dep[-1] = -1;
	
	for (int i = 1; i <= m; i++)
	{
		int a, b;
		scanf("%d%d", &a, &b);

		int u = (dep[a] > dep[b] ? a : b);  // u 较深
		int v = (dep[a] > dep[b] ? b : a);  // v 较浅

		if (dep[u] != dep[v])
		{
			for (int j = 20; j >= 0; j--)  // 使 u v 深度相同
			{
				int t = fa[u][j];
				if (dep[t] == dep[v])
				{
					u = t;
					break;
				}
				else if (dep[t] < dep[v])
					continue;
				u = t;
			}
		}

		if (u == v)
		{
			printf("%d\n", u);
			continue;
		}
		
		for (int j = 20; j >= 0; j--)  // 两点一起向上找公共父节点
		{
			int p = fa[u][j];
			int q = fa[v][j];
			if (p == q)
				continue;
			u = p;
			v = q;
		}
		printf("%d\n",fa[u][0]);
	}
	return 0;
}