LG11755

树链剖分模板题。边权转点权的操作十分经典，用两个点中深度较大的存储边权（因为父亲唯一），然后就转化为点权的操作了。轻度卡常即可通过，比如线段树查询时写非递归版等。

时间复杂度 \(O(n \log n + q \log^2 n )\)。

#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
#define N 2000010

using namespace std;

vector<int> G[N];
int dep[N],top[N],fa[N],siz[N],son[N],id[N],rev[N],ans1[N],ans2[N],eid[N],tot;
struct E
{
	int u,v;
}e[N];

int n,q;

struct T
{
	int t[N * 4];
	void update( int u , int l , int r , int L , int R , int x )
	{
		if( L > R ) return;
		if( L <= l && r <= R )
		{
			t[u] = x;
			return;
		}
		if( t[u] )
		{
			t[u << 1] = t[u];
			t[u << 1 | 1] = t[u];
			t[u] = 0;
		}
		int mid = ( l + r ) >> 1;
		if( L <= mid ) update( u << 1 , l , mid , L , R , x );
		if( R > mid ) update( u << 1 | 1 , mid + 1 , r , L , R , x );
		return;
	}
	int que( int p )
	{
		int l = 1,r = n,u = 1,mid;
		while( l < r )
		{
			mid = ( l + r ) >> 1;
			if( t[u] )
			{
				t[u << 1] = t[u];
				t[u << 1 | 1] = t[u];
				t[u] = 0;
			}
			if( p <= mid ) r = mid,u = u << 1;
			else l = mid + 1,u = u << 1 | 1;
		}
		return t[u];
	}
}tr;

void upd( int u , int v , int x )
{
	while( top[u] != top[v] )
	{	
		if( dep[top[u]] > dep[top[v]] ) swap( u , v );
		tr.update( 1 , 1 , n , id[top[v]] , id[v] , x );
		v = fa[top[v]];
	}
	if( dep[u] > dep[v] ) swap( u , v );
	tr.update( 1 , 1 , n , id[u] + 1 , id[v] , x );//由于是边权，故 LCA 不属于修改范围
	return;
}

void dfs1( int u , int f )
{
	fa[u] = f;
	dep[u] = dep[f] + 1;
	siz[u] = 1;
	for( auto v : G[u] )
	{
		if( v == f ) continue;
		dfs1( v , u );
		siz[u] += siz[v];
		if( siz[v] > siz[son[u]] )
			son[u] = v;
	}
	return; 
}

void dfs2( int u , int f , int flag )
{
	id[u] = ++ tot;
	rev[tot] = u;
	if( flag ) top[u] = u;
	else top[u] = top[f];
	if( son[u] )
		dfs2( son[u] , u , 0 );
	for( auto v : G[u] )
	{
		if( v == f || v == son[u] ) continue;
		dfs2( v , u , 1 );
	}
	return;
}

int read( void )
{
	int x = 0;
	char ch = getchar();
	while( ch < '0' || ch > '9' ) ch = getchar();
	while( ch >= '0' && ch <= '9' ) x = x * 10 + ch - '0',ch = getchar();
	return x; 
}

void write( int x )
{
	if( x <= 9 )
	{
		putchar( x + '0' );
		return;
	}
	write( x / 10 );
	putchar( x % 10 + '0' );
	return;
}

int main()
{
	int u,v;
	cin >> n >> q;
	for( int i = 1 ; i < n ; i ++ )
	{
		u = read(),v = read();
		e[i].u = u,e[i].v = v;
		G[u].push_back( v );
		G[v].push_back( u );
	}
	dfs1( 1 , 0 );
	dfs2( 1 , 0 , 1 );
	for( int i = 1 ; i <= q ; i ++ )
	{
		u = read(),v = read();
		upd( u , v , i );
	}
	for( int i = 1 ; i <= n ; i ++ )
		ans1[i] = tr.que( id[i] );
	for( int i = 1 ; i < n ; i ++ )
	{
		if( dep[e[i].u] < dep[e[i].v] ) swap( e[i].u , e[i].v );
		write( ans1[e[i].u] ),putchar( ' ' );
	}
	return 0;
}