Water Tree(树链剖分+dfs时间戳)

 Water Tree

http://codeforces.com/problemset/problem/343/D

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 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

 

树链剖分模板题

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cmath>
  5 #include<cstdio>
  6 #include<algorithm>
  7 #include<vector>
  8 #define maxn 500005
  9 #define lson l,mid,rt<<1
 10 #define rson mid+1,r,rt<<1|1
 11 using namespace std;
 12 
 13 int tree[maxn<<3],lazy[maxn<<3];
 14 int n;
 15 int dep[maxn],fa[maxn],siz[maxn],son[maxn],id[maxn],top[maxn],cnt;
 16 vector<int>ve[maxn];
 17 
 18 void pushup(int rt){
 19     if(tree[rt<<1]||tree[rt<<1|1])
 20         tree[rt]=1;
 21     else
 22         tree[rt]=0;
 23 }
 24 
 25 void pushdown(int rt){
 26     if(lazy[rt]!=-1){
 27         lazy[rt<<1]=lazy[rt];
 28         lazy[rt<<1|1]=lazy[rt];
 29         tree[rt<<1]=lazy[rt];
 30         tree[rt<<1|1]=lazy[rt];
 31         lazy[rt]=-1;
 32     }
 33 }
 34 
 35 void build(int l,int r,int rt){
 36     lazy[rt]=-1;
 37     if(l==r){
 38         tree[rt]=0;
 39         return;
 40     }
 41     int mid=(l+r)/2;
 42     build(lson);
 43     build(rson);
 44     pushup(rt);
 45 }
 46 
 47 void add(int L,int R,int k,int l,int r,int rt){
 48     if(L<=l&&R>=r){
 49         tree[rt]=k;
 50         lazy[rt]=k;
 51         return;
 52     }
 53     int mid=(l+r)/2;
 54     pushdown(rt);
 55     if(L<=mid) add(L,R,k,lson);
 56     if(R>mid) add(L,R,k,rson);
 57     pushup(rt);
 58 }
 59 
 60 int query(int L,int R,int l,int r,int rt){
 61     if(L<=l&&R>=r){
 62         return tree[rt];
 63     }
 64     int mid=(l+r)/2;
 65     pushdown(rt);
 66     int ans=0;
 67     if(L<=mid) if(ans||query(L,R,lson)) ans=1;
 68     if(R>mid) if(ans||query(L,R,rson)) ans=1;
 69     pushup(rt);
 70     return ans;
 71 }
 72 
 73 void dfs1(int now,int f,int deep){
 74     dep[now]=deep;
 75     siz[now]=1;
 76     fa[now]=f;
 77     int maxson=-1;
 78     for(int i=0;i<ve[now].size();i++){
 79         if(ve[now][i]==f) continue;
 80         dfs1(ve[now][i],now,deep+1);
 81         siz[now]+=siz[ve[now][i]];
 82         if(siz[ve[now][i]]>maxson){
 83             maxson=siz[ve[now][i]];
 84             son[now]=ve[now][i];
 85         }
 86     }
 87 }
 88 
 89 void dfs2(int now,int topp){
 90     id[now]=++cnt;
 91     top[now]=topp;
 92     if(!son[now]) return;
 93     dfs2(son[now],topp);
 94     for(int i=0;i<ve[now].size();i++){
 95         if(ve[now][i]==son[now]||ve[now][i]==fa[now]) continue;
 96         dfs2(ve[now][i],ve[now][i]);
 97     }
 98 }
 99 
100 void addRange(int x,int y,int k){
101     while(top[x]!=top[y]){
102         if(dep[top[x]]<dep[top[y]]) swap(x,y);
103         add(id[top[x]],id[x],k,1,n,1);
104         x=fa[top[x]];
105     }
106     if(dep[x]>dep[y]) swap(x,y);
107     add(id[x],id[y],k,1,n,1);
108 }
109 
110 void addSon(int x,int k){
111     add(id[x],id[x]+siz[x]-1,k,1,n,1);
112 }
113 
114 int main(){
115     std::ios::sync_with_stdio(false);
116     cin>>n;
117     int m;
118     int pos,z,x,y;
119     for(int i=1;i<n;i++){
120         cin>>x>>y;
121         ve[x].push_back(y);
122         ve[y].push_back(x);
123     }
124     cnt=0;
125     dfs1(1,0,1);
126     dfs2(1,1);
127     build(1,n,1);
128     cin>>m;
129     for(int i=1;i<=m;i++){
130         cin>>pos>>x;
131         if(pos==1){
132             addSon(x,1);
133         }
134         else if(pos==2){
135             addRange(1,x,0);
136             add(id[x],id[x],0,1,n,1);
137         }
138         else if(pos==3){
139             if(query(id[x],id[x],1,n,1)){
140                 cout<<1<<endl;
141             }
142             else{
143                 cout<<0<<endl;
144             }
145         }
146     }
147 
148 }
View Code

 

posted on 2018-09-26 00:05  Fighting_sh  阅读(713)  评论(0)  编辑  收藏  举报

导航