D. Water Tree

D. 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:

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 a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — 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 c_i (1 ≤ c_i ≤ 3), v_i (1 ≤ v_i ≤ n), where c_i is the operation type (according to the numbering given in the statement), and v_i 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
思路：dfs序+线段树;
先dfs将点映射成线性的，然后维护两棵线段树，一个是往里面灌水的时间，一个是抽水的时间，抽水的时候是单点往上更新，因为这个点的

如果是某个点的字节点，那么查询这个这个父亲节点点所包含的子树时一点包含要更新的点，然后每个节点的灌水和抽水的最晚的时间就可以了，
然后两次查询比较时间先后就可以了。
复杂度n×log（n）;

  1 #include<stdio.h>
  2 #include<math.h>
  3 #include<queue>
  4 #include<algorithm>
  5 #include<string.h>
  6 #include<iostream>
  7 #include<stack>
  8 #include<vector>
  9 using namespace std;
 10 typedef long long LL;
 11 typedef vector<int>Ve;
 12 vector<Ve>vec(600000);
 13 int ans[600000];
 14 int id[600000];
 15 bool flag[600000];
 16 int cn = 0;
 17 int l[600000];
 18 int r[600000];
 19 int pre[600000];
 20 void dfs(int n);
 21 int tree1[600000*4];
 22 int tree0[600000*4];
 23 int query(int l,int r,int k,int nn,int mm,int *tr,int c);
 24 void update(int l,int r,int k,int nn,int mm,int *tr,int i,int c);
 25 int main (void)
 26 {
 27     int n,x,y;
 28     scanf("%d",&n);
 29     for(int i = 0; i < n-1; i++)
 30     {
 31         scanf("%d %d",&x,&y);
 32         vec[x].push_back(y);
 33         vec[y].push_back(x);
 34     }
 35     dfs(1);
 36     for(int i = 1; i <= n; i++)
 37         id[ans[i]] = i;
 38     int m;
 39     scanf("%d",&m);
 40     int ccn = 0;
 41     while(m--)
 42     {
 43         int val,c;
 44         ++ccn;
 45         scanf("%d %d",&val,&c);
 46         if(val == 1)
 47         {
 48             update(l[c],r[c],0,1,cn,tree1,ccn,1);
 49             //printf("%d\n",ccn);
 50         }
 51         else if(val == 2)
 52         {
 53             update(id[c],id[c],0,1,cn,tree0,ccn,0);
 54         }
 55         else
 56         {
 57             int a = query(id[c],id[c],0,1,cn,tree1,1);
 58             int b = query(l[c],r[c],0,1,cn,tree0,0);
 59             //printf("%d %d\n",a,b);
 60             printf("%d\n",a>b);
 61         }
 62     }
 63     return 0;
 64 }
 65 void dfs(int n)
 66 {
 67     flag[n] = true;
 68     ans[++cn] = n;
 69     l[n] = cn;
 70     for(int i = 0; i < vec[n].size(); i++)
 71     {
 72         int ic = vec[n][i];
 73         if(!flag[ic])
 74             pre[ic] = n,dfs(ic);
 75     }
 76     r[n] = cn;
 77 }
 78 void update(int l,int r,int k,int nn,int mm,int *tr,int i,int c)
 79 {
 80     if(l > mm||r < nn)
 81     {
 82         return ;
 83     }
 84     else if(l <= nn&&r >= mm)
 85     {
 86         tr[k] = max(tr[k],i);
 87     }
 88     else
 89     {
 90         if(tr[k]&&c == 1)tr[2*k+1] = tr[k],tr[2*k+2] = tr[k],tr[k] = 0;
 91         update(l,r,2*k+1,nn,(nn+mm)/2,tr,i,c);
 92         update(l,r,2*k+2,(nn+mm)/2+1,mm,tr,i,c);
 93         if(c == 0)
 94             tr[k] = max(tr[k],tr[2*k+1]),tr[k] = max(tr[k],tr[2*k+2]);
 95     }
 96 }
 97 int query(int l,int r,int k,int nn,int mm,int *tr,int c)
 98 {
 99     if(l > mm||r < nn)
100     {
101         return 0;
102     }
103     else if(l <= nn&&r >= mm)
104     {
105         return tr[k];
106     }
107     else
108     {   if(c&&tr[k])tr[2*k+1] = tr[k],tr[2*k+2] = tr[k],tr[k] = 0;
109         int nx = query(l,r,2*k+1,nn,(nn+mm)/2,tr,c);
110         int ny = query(l,r,2*k+2,(nn+mm)/2+1,mm,tr,c);
111         return max(nx,ny);
112     }
113 }