codeforces contest 343 problem D(线段树+dfs序)

 Water Tree

time limit per test
4 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 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：两种操作，将节点u置1 它的子树为1 ，将u置0 它的祖先全为0；

解：这题操作太复杂，所以需要转换思维，如果将一个节点置1 先查询它的子树节点是否有0 如果有就将它的父亲置0  然后直接将它的子树区间置1，如果将一个节点置0 将这个节点的左区间端点置0 这样查询时 查询这个节点的子树中 是否有为0的节点就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6+10;
typedef long long LL;
vector<int>p
;
struct node
{
int l, r, v, f;
}g
;
int cnt;
void dfs(int u,int f)
{
g[u].l=++cnt,g[u].f=f;
for(int i=0;i<p[u].size();i++)
{
int v=p[u][i];
if(v==f) continue;
dfs(v,u);
}
g[u].r=++cnt;
return ;
}
int sum[N<<1];
void build(int l,int r,int rt)
{
sum[rt]=0;
if(l==r) return ;
int mid=(l+r)/2;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
return ;
}
void pushup(int rt)
{
sum[rt]=min(sum[rt<<1],sum[rt<<1|1]);
return ;
}
void pushdown(int rt)
{
if(sum[rt]==1) sum[rt<<1]=sum[rt<<1|1]=1;
return ;
}
void del(int x,int l,int r,int rt)
{
if(l==r)
{
sum[rt]=0;
return ;
}
pushdown(rt);
int mid=(l+r)/2;
if(x<=mid) del(x,l,mid,rt<<1);
else del(x,mid+1,r,rt<<1|1);
pushup(rt);
return ;
}

int query(int L,int R,int l,int r,int rt)
{
if(l>=L&&r<=R) return sum[rt];
pushdown(rt);
int mid=(l+r)/2,x=1;
if(L<=mid) x=min(x,query(L,R,l,mid,rt<<1));
if(R>mid) x=min(x,query(L,R,mid+1,r,rt<<1|1));
pushup(rt);
return x;
}
void update(int L,int R,int l,int r,int rt)
{
if(l>=L&&r<=R)
{
sum[rt]=1;
return ;
}
pushdown(rt);
int mid=(l+r)/2,x=1;
if(L<=mid) update(L,R,l,mid,rt<<1);
if(R>mid) update(L,R,mid+1,r,rt<<1|1);
pushup(rt);
return ;
}

int main()
{
int n, q, x, y;
scanf("%d", &n);
for(int i=1;i<n;i++)
{

scanf("%d %d", &x, &y);
p[x].push_back(y),p[y].push_back(x);
}
cnt=0;
dfs(1,-1);
build(1,cnt,1);
scanf("%d", &q);
while(q--)
{
scanf("%d %d", &x, &y);
if(x==1)
{
if(g[y].f!=-1&&query(g[y].l,g[y].r,1,cnt,1)==0) del(g[g[y].f].l,1,cnt,1);
update(g[y].l,g[y].r,1,cnt,1);
}
else if(x==2) del(g[y].l,1,cnt,1);
else printf("%d\n",query(g[y].l,g[y].r,1,cnt,1));
}
return 0;
}