Codeforces-343D Water Tree（树链剖分）

Description：

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:

1. Fill vertex v with water. Then v and all its children are filled with water.
2. Empty vertex v. Then v and all its ancestors are emptied.
3. 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 1 2 3 3 1 3 2 3 3 3 4 1 2 2 4 3 1 3 3 3 4 3 5

output

0 0 0 1 0 1 0 1

题意：

给出一个具有N个点的树，1号节点是根节点。现在有三种操作：

1 x，表示给x节点灌溉水，并且其儿子都会被灌上水。

2 x，表示将x节点的水抽干，并且其父亲节点会一路被抽干。

3 x，表示询问x节点是否有水。有水输出1，否则输出0。

对于操作1：DFS的时候可以记录从每个点进入的时刻L[x]和最后出来的时刻R[X]，更新就是[L[X],R[X]]；对于操作2：就利用树链剖分从节点x向节点1靠拢并更新。

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<math.h>
#include<vector>
using namespace std;
typedef long long ll;
typedef double ld;
const int MAX=6e5;
vector<int>e[MAX];
struct lenka
{
    int l,r;
    int ans,tag;
}A[MAX<<2];
int d[MAX],son[MAX],fa[MAX],siz[MAX],top[MAX],L[MAX],R[MAX],all;
void build(int k,int l,int r)
{
    A[k].l=l,A[k].r=r;
    A[k].ans=A[k].tag=0;
    if(l==r)return;
    build(2*k,l,(l+r)/2);
    build(2*k+1,(l+r)/2+1,r);
}
void change(int k,int x,int y,int tag)
{
    if(x==A[k].l&&y==A[k].r)
    {
        A[k].ans=tag;
        if(x!=y)A[k].tag=1;
        return;
    }
    if(A[k].tag)
    {
        change(2*k,A[2*k].l,A[2*k].r,A[k].ans);
        change(2*k+1,A[2*k+1].l,A[2*k+1].r,A[k].ans);
        A[k].tag=0;
    }
    if(y<=A[2*k].r)change(2*k,x,y,tag);
    else if(x>=A[2*k+1].l)change(2*k+1,x,y,tag);
    else
    {
        change(2*k,x,A[2*k].r,tag);
        change(2*k+1,A[2*k+1].l,y,tag);
    }
}
int ask(int k,int x)
{
    if(x==A[k].l&&x==A[k].r)
        return A[k].ans;
    if(A[k].tag)
    {
        change(2*k,A[2*k].l,A[2*k].r,A[k].ans);
        change(2*k+1,A[2*k+1].l,A[2*k+1].r,A[k].ans);
        A[k].tag=0;
    }
    if(x<=A[2*k].r)
    return ask(2*k,x);
    return ask(2*k+1,x);
}
void dfs1(int k,int f,int dep)
{
    d[k]=dep;
    siz[k]=1;
    son[k]=0;
    fa[k]=f;
    for(int i=0;i<e[k].size();i++)
    {
        int nex=e[k][i];
        if(nex==f)continue;
        dfs1(nex,k,dep+1);
        siz[k]+=siz[nex];
        if(siz[son[k]]<siz[nex])son[k]=nex;
    }
}
void dfs2(int k,int tp)
{
    top[k]=tp;
    L[k]=++all;
    if(son[k])dfs2(son[k],tp);
    for(int i=0;i<e[k].size();i++)
    {
        if(e[k][i]==fa[k]||e[k][i]==son[k])
            continue;
        dfs2(e[k][i],e[k][i]);
    }
    R[k]=all;
}
void slove(int x,int y)
{
    while(top[x]!=top[y])
    {
        if(d[top[x]]<d[top[y]])swap(x,y);
        change(1,L[top[x]],L[x],0);
        x=fa[top[x]];
    }
    if(d[x]>d[y])swap(x,y);
    change(1,L[x],L[y],0);

}
int main()
{
    int n,m;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    all=0;
    dfs1(1,0,1);
    dfs2(1,1);
    build(1,1,all);
    scanf("%d",&m);
    while(m--)
    {
        int op,x;
        scanf("%d%d",&op,&x);
        if(op==1)
            change(1,L[x],R[x],1);
        else if(op==2)
        slove(1,x);
        else
        printf("%d\n",ask(1,L[x]));
    }
    return 0;
}