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.
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.
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
0
0
0
1
0
1
0
1
1 #include<iostream> 2 #include<stdio.h> 3 #include<string.h> 4 #include<algorithm> 5 using namespace std; 6 const int N=5e5+10; 7 int sum[N*4],lazy[N*4];//线段树 8 int n,m,r,mod;//节点数，操作数，根节点，模数 9 int first[N],tot; //邻接表 10 //重儿子，每个节点新编号，父亲，编号，深度，子树个数，所在重链的顶部 11 int son[N],id[N],fa[N],cnt,deep[N],size[N],top[N]; 12 int w[N],wt[N];// 初始点权，新编号点权 13 int res=0;//查询答案 14 15 struct edge{ 16 int v,next; 17 }e[N*4]; 18 19 void add_edge(int u,int v){ 20 e[tot].v=v; 21 e[tot].next=first[u]; 22 first[u]=tot++; 23 } 24 25 void init(){ 26 memset(first,-1,sizeof(first)); 27 tot=0; 28 cnt=0; 29 } 30 31 int pushup(int rt){ 32 sum[rt]=(sum[rt*2]&&sum[rt*2+1]); 33 } 34 35 void pushdown(int rt,int m){ 36 if(lazy[rt]!=-1){ 37 lazy[rt*2]=lazy[rt]; 38 lazy[rt*2+1]=lazy[rt]; 39 sum[rt*2]=lazy[rt]; 40 sum[rt*2+1]=lazy[rt]; 41 lazy[rt]=-1; 42 } 43 } 44 45 void build(int l,int r,int v){ 46 lazy[v]=-1; 47 if(l==r){ 48 sum[v]=0; 49 return ; 50 } 51 int mid=(l+r)/2; 52 build(l,mid,v*2); 53 build(mid+1,r,v*2+1); 54 pushup(v); 55 } 56 57 void update(int L,int R,int c,int l,int r,int rt){ 58 if(L<=l&&r<=R){ 59 lazy[rt]=c; 60 sum[rt]=c; 61 return; 62 } 63 pushdown(rt,r-l+1); 64 int m=(l+r)/2; 65 if(L<=m) update(L,R,c,l,m,rt*2); 66 if(R>m) update(L,R,c,m+1,r,rt*2+1); 67 pushup(rt); 68 } 69 70 71 void dfs1(int u,int f,int d){ 72 deep[u]=d; 73 fa[u]=f; 74 size[u]=1; 75 int maxson=-1; 76 for(int i=first[u];~i;i=e[i].next){ 77 int v=e[i].v; 78 if(v==f) continue; 79 dfs1(v,u,d+1); 80 size[u]+=size[v]; 81 if(size[v]>maxson){ 82 son[u]=v; 83 maxson=size[v]; 84 } 85 } 86 } 87 88 void dfs2(int u,int topf){ 89 id[u]=++cnt; 90 wt[cnt]=w[u]; 91 top[u]=topf; 92 if(!son[u]) return ; 93 dfs2(son[u],topf); 94 for(int i=first[u];~i;i=e[i].next){ 95 int v=e[i].v; 96 if(v==fa[u]||v==son[u]) continue; 97 dfs2(v,v); 98 } 99 } 100 101 102 void updrange(int x,int y,int k){ 103 while(top[x]!=top[y]){ 104 if(deep[top[x]]<deep[top[y]]) swap(x,y); 105 update(id[top[x]],id[x],0,1,n,1); 106 x=fa[top[x]]; 107 } 108 if(deep[x]>deep[y]) swap(x,y); 109 update(id[x],id[y],0,1,n,1); 110 } 111 112 void upson(int x,int k){ 113 update(id[x],id[x]+size[x]-1,1,1,n,1); 114 } 115 116 int query(int L,int R,int l,int r,int rt){ 117 if(L<=l&&r<=R){ 118 return sum[rt]; 119 } 120 pushdown(rt,r-l+1); 121 int m=(l+r)/2; 122 if(L<=m) return query(L,R,l,m,rt*2); 123 else if(R>m) return query(L,R,m+1,r,rt*2+1); 124 } 125 126 127 128 129 130 131 int main(){ 132 int u,v; 133 scanf("%d",&n); 134 init(); 135 for(int i=1;i<=n-1;i++){ 136 scanf("%d%d",&u,&v); 137 add_edge(u,v); 138 add_edge(v,u); 139 } 140 dfs1(1,0,1); 141 dfs2(1,1); 142 build(1,n,1); 143 scanf("%d",&m); 144 while(m--){ 145 int op,x,y,z; 146 scanf("%d",&op); 147 if(op==1){ 148 scanf("%d",&x); 149 upson(x,1); 150 } 151 else if(op==2){ 152 scanf("%d",&x); 153 updrange(1,x,0); 154 } 155 else if(op==3){ 156 scanf("%d",&x); 157 if(query(id[x],id[x],1,n,1)==0) printf("0\n"); 158 else printf("1\n"); 159 } 160 } 161 }
Tree will split the chain, this question is the dimensionality reduction blow