Description
Mad scientist Mike has constructed a rooted tree, which consists of nnvertices. Each vertex is a reservoir which can be either empty or filled with water.
The vertices of the tree are numbered from 1 to nn 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 vv with water. Then vv and all its children are filled with water.
Empty vertex vv . Then vv and all its ancestors are emptied.
Determine whether vertex vv 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-1n−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.
你不需要理解题意,你只需要知道,这是一个树剖裸题(虽然我没一遍切。)
支持三种操作(初始值全部为\(0\))
- 1.将节点\(v\)及其子树赋值为\(1\).
- 2.将节点\(v\)到根节点\(1\)的路径上的点的值置为\(0\).
- 3.查询当前节点\(v\)的值。(只会为\(0\)或\(1\))
对于每个操作\(3\),输出一行。(具体见代码好了
这是一个不完整的树剖,我没建树,有没用到反\(dfs\)序。 emmm
代码
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define R register
using namespace std;
const int gz=500001;
inline void in(int &x)
{
int f=1;x=0;char s=getchar();
while(!isdigit(s)){if(s=='-')f=-1;s=getchar();}
while(isdigit(s)){x=x*10+s-'0';s=getchar();}
x*=f;
}
int head[gz],tot,n,m;
struct cod{int u,v;}edge[gz<<1];
inline void add(R int x,R int y)
{
edge[++tot].u=head[x];
edge[tot].v=y;
head[x]=tot;
}
int dfn[gz],idx,son[gz],f[gz],depth[gz],size[gz],top[gz];
void dfs1(R int u,R int fa)
{
f[u]=fa;depth[u]=depth[fa]+1;size[u]=1;
for(R int i=head[u];i;i=edge[i].u)
{
if(edge[i].v==fa)continue;
dfs1(edge[i].v,u);
size[u]+=size[edge[i].v];
if(son[u]==-1 or size[son[u]]<size[edge[i].v])
son[u]=edge[i].v;
}
}
void dfs2(R int u,R int t)
{
dfn[u]=++idx;top[u]=t;
if(son[u]==-1)return ;
dfs2(son[u],t);
for(R int i=head[u];i;i=edge[i].u)
{
if(dfn[edge[i].v])continue;
dfs2(edge[i].v,edge[i].v);
}
}
int tg[gz<<2],tr[gz<<2];
#define ls o<<1
#define rs o<<1|1
inline void down(R int o)
{
if(tg[o]==-1)return;
tg[ls]=tg[rs]=tg[o];
tr[ls]=tr[rs]=tr[o];
tg[o]=-1;
return ;
}
void change(R int o,R int l,R int r,R int x,R int y,R int k)
{
if(x<=l and y>=r){tr[o]=tg[o]=k;return;}
down(o);
R int mid=(l+r)>>1;
if(x<=mid)change(ls,l,mid,x,y,k);
if(y>mid)change(rs,mid+1,r,x,y,k);
}
int query(R int o,R int l,R int r,R int pos)
{
if(l==r)return tr[o];
down(o);
R int mid=(l+r)>>1;
if(pos<=mid)return query(ls,l,mid,pos);
else return query(rs,mid+1,r,pos);
}
inline void tchange(R int x,R int y)
{
R int fx=top[x],fy=top[y];
while(fx!=fy)
{
if(depth[fx]>depth[fy])
{
change(1,1,n,dfn[fx],dfn[x],0);
x=f[fx];
}
else
{
change(1,1,n,dfn[fy],dfn[y],0);
y=f[fy];
}
fy=top[y],fx=top[x];
}
if(dfn[x]>dfn[y])swap(x,y);
change(1,1,n,dfn[x],dfn[y],0);
return ;
}
int main()
{
in(n);memset(son,-1,sizeof son);
for(R int i=1,x,y;i<n;i++)
{
in(x),in(y);
add(x,y),add(y,x);
}
dfs1(1,0);dfs2(1,1);memset(tg,-1,sizeof tg);
in(m);
for(R int i=1,opt,v;i<=m;i++)
{
in(opt);
switch(opt)
{
case 1:in(v);change(1,1,n,dfn[v],dfn[v]+size[v]-1,1);break;
case 2:in(v);tchange(1,v);break;
case 3:in(v);printf("%d\n",query(1,1,n,dfn[v]));break;
}
}
}