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:

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.

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.

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.

51 25 12 34 2121 12 33 13 23 33 41 22 43 13 33 43 5output
00010101
題解：樹鏈剖分
小水題，不解釋。。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define N 1000003using namespace std;int n,m,tot,l[N],r[N],point[N],nxt[N],v[N],sz;int fa[N],deep[N],size[N],son[N],pos[N],belong[N],tr[N*4],delta[N*4];void add(int x,int y){	tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y;	tot++; nxt[tot]=point[y]; point[y]=tot; v[tot]=x;}void dfs(int x,int f){	size[x]=1; deep[x]=deep[f]+1;	for (int i=point[x];i;i=nxt[i]) {		if (v[i]==f) continue;		fa[v[i]]=x;		dfs(v[i],x);		size[x]+=size[v[i]];		if (size[v[i]]>size[son[x]]) son[x]=v[i];	}}void dfs1(int x,int chain){	belong[x]=chain; pos[x]=++sz; l[x]=r[x]=sz;	if (!son[x]) return ;	dfs1(son[x],chain);	for (int i=point[x];i;i=nxt[i])	 if(v[i]!=fa[x]&&v[i]!=son[x]) dfs1(v[i],v[i]);	r[x]=sz;}void pushdown(int now){	if (delta[now]!=-1) {		delta[now<<1]=delta[now];		delta[now<<1|1]=delta[now];		tr[now<<1]=delta[now]; tr[now<<1|1]=delta[now];		delta[now]=-1;	}}void qjchange(int now,int l,int r,int ll,int rr,int val){	if (ll<=l&&r<=rr) {		tr[now]=val;		delta[now]=val;		return;	}	int mid=(l+r)/2;	pushdown(now);	if (ll<=mid)  qjchange(now<<1,l,mid,ll,rr,val);	if (rr>mid) qjchange(now<<1|1,mid+1,r,ll,rr,val);}int find(int now,int l,int r,int x){	if (l==r) return tr[now];	int mid=(l+r)/2;	pushdown(now);	if (x<=mid) return find(now<<1,l,mid,x);	else return find(now<<1|1,mid+1,r,x);}void solve(int x,int y){	while (belong[x]!=belong[y]){		if (deep[belong[x]]<deep[belong[y]]) swap(x,y);		qjchange(1,1,n,pos[belong[x]],pos[x],0);        x=fa[belong[x]];	}	if (deep[x]>deep[y]) swap(x,y);	qjchange(1,1,n,pos[x],pos[y],0);}int main(){	freopen("a.in","r",stdin);	scanf("%d",&n);	for (int i=1;i<n;i++) {		int x,y; scanf("%d%d",&x,&y);		add(x,y);	}	dfs(1,0); dfs1(1,1);	memset(delta,-1,sizeof(delta));	scanf("%d",&m);	for (int i=1;i<=m;i++) {		int opt,x; scanf("%d%d",&opt,&x);		if (opt==1) qjchange(1,1,n,l[x],r[x],1);		if (opt==2) solve(1,x);		if (opt==3) printf("%d/n",find(1,1,n,pos[x]));	}}