BZOJ3720 Gty的妹子树 询问分块、主席树

传送门

---

学到了询问分块的科技……

对于修改操作，每发生了\(S\)次修改就重构整棵树，小于\(S\)次的修改操作丢到一个队列里面。

对于每一次查询操作，先在主席树上查询当前子树内部大于\(k\)的节点的数量，然后依次将队列中的修改放到树上，在答案统计完成之后再将这些修改撤销。使用倍增检验某一个点是否在子树内，如果在子树内则考虑这个节点的权值修改或加入对于答案的影响。

修改的复杂度为\(O(\frac{N}{S}NlogN)\)，查询的复杂度为\(O(NSlogN)\)，当\(S = \sqrt{N}\)时有最优复杂度\(O(N \sqrt N logN)\)

因为分块的玄学性，\(S\)取\(\sqrt N\)可能会T，取\(\sqrt N logN\)跑得挺快。。。

#include<bits/stdc++.h>
#define mid ((l + r) >> 1)
#define st first
#define nd second
#define PII pair < int , int >
#define PIII pair < int , pair < int , int > >
//This code is written by Itst
using namespace std;

inline int read(){
	int a = 0;
	char c = getchar();
	while(!isdigit(c))
		c = getchar();
	while(isdigit(c)){
		a = a * 10 + c - 48;
		c = getchar();
	}
	return a;
}

const int MAXN = 6e4 + 7;
struct Edge{
	int end , upEd;
}Ed[MAXN << 1];
struct node{
	int l , r , sum;
}Tree[MAXN * 30];
int head[MAXN] , val[MAXN] , dfn[MAXN] , sz[MAXN] , dep[MAXN] , jump[MAXN][21];
int N , N1 , M , T , ts , cntN , cntEd , cntL , lsh[MAXN] , rt[MAXN];
deque < PIII > q;
queue < PII > rev;
bool vis[MAXN];

inline void addEd(int a , int b){
	Ed[++cntEd].end = b;
	Ed[cntEd].upEd = head[a];
	head[a] = cntEd;
}

int insert(int x , int l , int r , int tar){
	int t = ++cntN;
	Tree[t] = Tree[x];
	++Tree[t].sum;
	if(l != r)
		if(mid >= tar)
			Tree[t].l = insert(Tree[t].l , l , mid , tar);
		else
			Tree[t].r = insert(Tree[t].r , mid + 1 , r , tar);
	return t;
}

void dfs(int x , int p){
	dep[x] = dep[p] + 1;
	dfn[x] = ++ts;
	sz[x] = 1;
	rt[ts] = insert(rt[ts - 1] , 1 , cntL , val[x]);
	jump[x][0] = p;
	for(int i = 1 ; jump[x][i - 1] ; ++i)
		jump[x][i] = jump[jump[x][i - 1]][i - 1];
	for(int i = head[x] ; i ; i = Ed[i].upEd)
		if(Ed[i].end != p){
			dfs(Ed[i].end , x);
			sz[x] += sz[Ed[i].end];
		}
}

inline void build(){
	for(int i = 1 ; i <= N ; ++i)
		lsh[i] = val[i];
	sort(lsh + 1 , lsh + N + 1);
	cntL = unique(lsh + 1 , lsh + N + 1) - lsh - 1;
	for(int i = 1 ; i <= N ; ++i)
		val[i] = lower_bound(lsh + 1 , lsh + cntL + 1 , val[i]) - lsh;
	dfs(1 , 0);
}

inline void rebuild(){
	cntN = ts = 0;
	for(int i = 1 ; i <= N ; ++i)
		val[i] = lsh[val[i]];
	while(!q.empty()){
		PIII t = q.front();
		q.pop_front();
		if(t.st == 1)
			val[t.nd.st] = t.nd.nd;
		else
			addEd(jump[t.nd.st][0] , t.nd.st);
	}
	N = N1;
	build();
}

int query(int x , int l , int r , int tar){
	if(l == r || !x)
		return 0;
	if(mid >= tar)
		return query(Tree[x].l , l , mid , tar) + Tree[Tree[x].r].sum;
	else
		return query(Tree[x].r , mid + 1 , r , tar);
}

inline int to(int x , int d){
	for(int i = 16 ; i >= 0 ; --i)
		if(dep[x] - (1 << i) >= d)
			x = jump[x][i];
	return x;
}

int get(int x , int v){
	int sum , l = q.size();
	if(v >= lsh[cntL])
		sum = 0;
	else
		if(v < lsh[1])
			sum = sz[x];
		else
			sum = query(rt[dfn[x] + sz[x] - 1] , 1 , cntL , upper_bound(lsh + 1 , lsh + cntL + 1 , v) - lsh - 1) - query(rt[dfn[x] - 1] , 1 , cntL , upper_bound(lsh + 1 , lsh + cntL + 1 , v) - lsh - 1);
	for(int i = 0 ; i < l ; ++i){
		int t = q[i].nd.st;
		if(dep[t] < dep[x] || to(t , dep[x]) != x)
			continue;
		if(q[i].st == 1){
			if(!vis[t]){
				vis[t] = 1;
				rev.push(PII(t , val[t]));
				if(t <= N)
					val[t] = lsh[val[t]];
			}
			if(val[t] <= v && q[i].nd.nd > v)
				++sum;
			else
				if(val[t] > v && q[i].nd.nd <= v)
					--sum;
			val[t] = q[i].nd.nd;
		}
		else
			if(val[t] > v)
				++sum;
	}
	while(!rev.empty()){
		PII t = rev.front();
		rev.pop();
		vis[t.st] = 0;
		val[t.st] = t.nd;
	}
	return sum;
}

int main(){
#ifndef ONLINE_JUDGE
	freopen("in","r",stdin);
	freopen("out","w",stdout);
#endif
	N1 = N = read();
	T = sqrt(N * log2(N));
	for(int i = 1 ; i < N ; ++i){
		int a = read() , b = read();
		addEd(a , b);
		addEd(b , a);
	}
	for(int i = 1 ; i <= N ; ++i)
		val[i] = read();
	build();
	M = read();
	int a , b , c , lastans = 0;
	for(int i = 1 ; i <= M ; ++i){
		a = read();
		b = read() ^ lastans;
		c = read() ^ lastans;
		if(!a){
			if(q.size() >= T)
				rebuild();
			printf("%d\n" , lastans = get(b , c));
		}
		else{
			if(a == 2){
				val[++N1] = c;
				dep[N1] = dep[b] + 1;
				jump[N1][0] = b;
				for(int j = 1 ; jump[N1][j - 1] ; ++j)
					jump[N1][j] = jump[jump[N1][j - 1]][j - 1];
			}
			q.push_back(PIII(a , PII(a == 2 ? N1 : b , c)));
		}
	}
	return 0;
}