CF343D Water Tree

題目描述

Mad scientist Mike has constructed a rooted tree, which consists of nn 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 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:

1. Fill vertex vv with water. Then vv and all its children are filled with water.
2. Empty vertex vv . Then vv and all its ancestors are emptied.
3. 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.

輸入格式

The first line of the input contains an integer nn ( 1<=n<=5000001<=n<=500000 ) — the number of vertices in the tree. Each of the following n-1n−1 lines contains two space-separated numbers a_{i}a**i , b_{i}b**i ( 1<=a_{i},b_{i}<=n1<=a**i,b**i<=n , a_{i}≠b_{i}a**i=b**i ) — the edges of the tree.

The next line contains a number qq ( 1<=q<=5000001<=q<=500000 ) — the number of operations to perform. Each of the following qq lines contains two space-separated numbers c_{i}c**i ( 1<=c_{i}<=31<=c**i<=3 ), v_{i}v**i ( 1<=v_{i}<=n1<=v**i<=n ), where c_{i}c**i is the operation type (according to the numbering given in the statement), and v_{i}v**i is the vertex on which the operation is performed.

It is guaranteed that the given graph is a tree.

輸出格式

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.

題意翻譯

• 給出一棵以 11 為根節點的 nn 個節點的有根樹。每個點有一個權值，初始為 00。

• mm

  次操作。操作有

  33

  種：

  1. 將點 uu 和其子樹上的所有節點的權值改為 11。
  2. 將點 uu 到 11 的路徑上的所有節點的權值改為 00。
  3. 詢問點 uu 的權值。

• 1\le n,m\le 5\times 10^51≤n,m≤5×105。

題解：

樹剖架線段樹。

線段樹要維護的操作是區間覆蓋和區間清空。

題解一堆珂朵莉樹？

不明白這種一點都不嚴肅的演算法名稱是哪個誰起的。

反正我用的是樹剖架線段樹

程式碼：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=1e6+7;
struct node{
	int l,r,sum,lazy;
}tree[maxn<<2];
int head[maxn],nxt[maxn<<1],to[maxn<<1],tot;
int n,m,x,y;
void add(int x,int y)
{
	nxt[++tot]=head[x];
	to[tot]=y;
	head[x]=tot;
}
int fa[maxn],deep[maxn],id[maxn],val[maxn],size[maxn],son[maxn],top[maxn],cnt;
void dfs1(int x,int f)
{
	fa[x]=f;
	size[x]=1;
	deep[x]=deep[f]+1;
	int maxson=-1;
	for(int i=head[x];i;i=nxt[i])
	{
		int y=to[i];
		if(y==f) 
			continue;
		dfs1(y,x);
		size[x]+=size[y];
		if(size[y]>maxson)
		{
			maxson=size[y];
			son[x]=y;
		}
	}
}
void dfs2(int x,int t)
{
	top[x]=t;
	id[x]=++cnt;
	if(!son[x]) 
		return;
	dfs2(son[x],t);
	for(int i=head[x];i;i=nxt[i])
	{
		int y=to[i];
		if(y==son[x]||y==fa[x]) 
			continue;
		dfs2(y,y);
	}
}
void build(int pos,int l,int r)
{
	tree[pos].l=l,tree[pos].r=r,tree[pos].lazy=-1;
	if(l==r) 
		return;
	int mid=(l+r)>>1;
	build(pos<<1,l,mid);
	build(pos<<1|1,mid+1,r);
}
void pushdown(int pos)
{
	if(tree[pos].lazy!=-1)
	{
		tree[pos<<1].sum=(tree[pos<<1].r-tree[pos<<1].l+1)*tree[pos].lazy;
		tree[pos<<1|1].sum=(tree[pos<<1|1].r-tree[pos<<1|1].l+1)*tree[pos].lazy;
		tree[pos<<1].lazy=tree[pos].lazy;
		tree[pos<<1|1].lazy=tree[pos].lazy;
		tree[pos].lazy=-1;
	}
}
void update(int pos,int l,int r,int y)
{
	if(tree[pos].l>=l&&tree[pos].r<=r)
	{
		tree[pos].sum=(tree[pos].r-tree[pos].l+1)*y;
		tree[pos].lazy=y;
		return;
	}
	pushdown(pos);
	int mid=(tree[pos].l+tree[pos].r)>>1;
	if(l<=mid) 
		update(pos<<1,l,r,y);
	if(r>mid) 
		update(pos<<1|1,l,r,y);
	tree[pos].sum=tree[pos<<1].sum+tree[pos<<1|1].sum;
}
int query(int pos,int l,int r)
{
	if(tree[pos].l>=l&&tree[pos].r<=r) 
		return tree[pos].sum;
	pushdown(pos);
	int mid=(tree[pos].l+tree[pos].r)>>1;
	int val=0;
	if(l<=mid) 
		val+=query(pos<<1,l,r);
	if(r>mid) 
		val+=query(pos<<1|1,l,r);
	return val; 
}
void upd_chain(int x,int y,int k)
{
	while(top[x]!=top[y])
	{
		if(deep[top[x]]<deep[top[y]]) 
			swap(x,y);
		update(1,id[top[x]],id[x],k);
		x=fa[top[x]];
	}
	if(deep[x]<deep[y]) 
		swap(x,y);
	update(1,id[y],id[x],k);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		add(x,y);
		add(y,x);
	}
	dfs1(1,0);
	dfs2(1,1);
	build(1,1,n);
	scanf("%d",&m);
	int opt,x;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&opt,&x);
		if(opt==1) 
			update(1,id[x],id[x]+size[x]-1,1); 
		else if(opt==2) 
			upd_chain(x,1,0);
		else 
			printf("%d\n",query(1,id[x],id[x]));
	}
	return 0;
}