Water Tree CodeForces - 343D(트리 체인 분할 + 세그먼트 트리)

Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Each vertex is a reservoir which can be either empty or filled with water.
The vertices of the tree are numbered from 1 to n with the root at vertex 1. For each vertex, the reservoirs of its children are located below the reservoir of this vertex, and the vertex is connected with each of the children by a pipe through which water can flow downwards.
Mike wants to do the following operations with the tree:
Fill vertex v with water. Then v and all its children are filled with water. Empty vertex v. Then v and all its ancestors are emptied. Determine whether vertex v is filled with water at the moment. Initially all vertices of the tree are empty. Mike has already compiled a full list of operations that he wants to perform in order. Before experimenting with the tree Mike decided to run the list through a simulation. Help Mike determine what results will he get after performing all the operations.
Input The first line of the input contains an integer n (1 ≤ n ≤ 500000) — the number of vertices in the tree. Each of the following n - 1 lines contains two space-separated numbers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the edges of the tree.
The next line contains a number q (1 ≤ q ≤ 500000) — the number of operations to perform. Each of the following q lines contains two space-separated numbers ci (1 ≤ ci ≤ 3), vi (1 ≤ vi ≤ n), where ci is the operation type (according to the numbering given in the statement), and vi is the vertex on which the operation is performed.
It is guaranteed that the given graph is a tree.
Output For each type 3 operation print 1 on a separate line if the vertex is full, and 0 if the vertex is empty. Print the answers to queries in the order in which the queries are given in the input.
Examples Input 5 1 2 5 1 2 3 4 2 12 1 2 3 3 3 3 3 4 1 2 2 4 2 4 3 3 3 3 3 4 3 5 Output 0 0 0 1 0 1 0 1 0 1 0 1 1 1 은 트리 체인을 분할한 템플릿 문제입니다. 온라인 세그먼트 트리를 조회하고 업데이트할 때 최적화를 선택할 수 있습니다. 최적화되지 않으면 통과할 수 있는지 모르겠습니다.처음으로 이런 문제를 혼자 풀어봤는데 호호호 코드는 다음과 같다.

#include
#include
#include
#include
#include
#include
#include
#define ll long long
using namespace std;

const int maxx=5e5+100;
int head[maxx<<1],pre[maxx],id[maxx];
int size[maxx],son[maxx],fa[maxx];
int dep[maxx],top[maxx];
struct edge{
	int to;
	int next;
}e[maxx<<1];
struct node{
	int l;
	int r;
	int lazy;
	int sum;
}p[maxx*4];
int n,m,tot,sign;
/*------------    --------------*/ 
void init()
{
	memset(head,-1,sizeof(head));
	tot=sign=0;
}
void add(int u,int v)
{
	e[tot].to=v,e[tot].next=head[u],head[u]=tot++;
}
/*--------------dfs-----------------*/
void dfs1(int u,int f)
{
	dep[u]=dep[f]+1;
	fa[u]=f;
	size[u]=1;
	for(int i=head[u];i!=-1;i=e[i].next)
	{
		int to=e[i].to;
		if(to==f) continue;
		dfs1(to,u);
		size[u]+=size[to];
		if(size[to]>size[son[u]]) son[u]=to;
	}
}
void dfs2(int u,int Top)
{
	id[u]=++sign;
	pre[sign]=u;
	top[u]=Top;
	if(son[u]) dfs2(son[u],Top);
	for(int i=head[u];i!=-1;i=e[i].next)
	{
		int to=e[i].to;
		if(to==son[u]||to==fa[u]) continue;
		dfs2(to,to);
	}
}
/*--------------   --------------*/
void pushup(int cur){p[cur].sum=p[cur<<1].sum+p[cur<<1|1].sum;}
void pushdown(int cur)
{
	if(p[cur].lazy>0)
	{
		p[cur<<1].sum=p[cur<<1].r-p[cur<<1].l+1;
		p[cur<<1|1].sum=p[cur<<1|1].r-p[cur<<1|1].l+1;
		p[cur<<1].lazy=1;
		p[cur<<1|1].lazy=1;
		p[cur].lazy=0;
	}
	else if(p[cur].lazy<0)
	{
		p[cur<<1].sum=p[cur<<1|1].sum=0;
		p[cur<<1].lazy=p[cur<<1|1].lazy=-1;
		p[cur].lazy=0;
	}
}
void build(int l,int r,int cur)
{
	p[cur].l=l;
	p[cur].r=r;
	p[cur].sum=0;
	if(l==r) return ;
	int mid=(l+r)/2;
	build(l,mid,2*cur);
	build(mid+1,r,2*cur+1);
	pushup(cur);
}
void update(int l,int r,int cur,int add)
{
	if(p[cur].sum==p[cur].r-p[cur].l+1&&add==1) return ;
	if(p[cur].sum==0&&add==-1) return ;//         ，              ，       
	int L=p[cur].l;
	int R=p[cur].r;
	if(l<=L&&R<=r)
	{
		if(add>0)
		{
			p[cur].sum=(R-L+1);
			p[cur].lazy=1;
		}
		else 
		{
			p[cur].sum=0;
			p[cur].lazy=-1;
		} 
		return ;
	}
	pushdown(cur);
	int mid=(L+R)/2;
	if(r<=mid) update(l,r,2*cur,add);
	else if(l>mid) update(l,r,2*cur+1,add);
	else 
	{
		update(l,mid,2*cur,add);
		update(mid+1,r,2*cur+1,add);
	}
	pushup(cur);
}
int query(int x,int cur)
{
	int L=p[cur].l;
	int R=p[cur].r;
	if(p[cur].sum==R-L+1) return 1；//   
	if(L==R)
	{
		if(p[cur].sum) return 1;
		else return 0;
	}
	pushdown(cur);
	int mid=(L+R)/2;
	if(x<=mid) return query(x,2*cur);
	else return query(x,2*cur+1);
}
/*------------    -------------*/
void solve(int x,int y)
{
	while(top[x]!=top[y])
	{
		if(dep[top[x]]id[y]) swap(x,y);
	update(id[x],id[y],1,-1);
} 
int main()
{
	init();
	scanf("%d",&n);
	int x,y;
	for(int i=1;i

열심히 힘내라 a야,(o)/~