CF-877-E-dfs序+线段树

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-237.CF-282-E-Trie2024-01-178.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-1210.CF-431-D-二分+数位DP2024-01-2411.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-16

22.CF-877-E-dfs序+线段树2024-05-05

23.CF-915-F-并查集2024-01-2124.CF-938-D-最短路2024-05-1025.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-2231.CF-1921-F-根号分治2024-01-24

收起

877-E 题目大意

给定一颗$n$个节点的树，根为$1$，点带权，权值要么为0，要么为1。

$q$次询问，两种类型：

• $getx$:询问$x$的子树中有多少个$1$。
• $powx$:将$x$子树中所有的值取反。

---

Solution

dfs序+线段树模板题，把子树上的操作转化为区间上的操作。

时间复杂度：$O(nlogn)$。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N=2e5+10;
struct node{
	int l,r;
	int sum,tag;
}tr[N<<2];

void pushup(int u){
	tr[u].sum=tr[u<<1].sum+tr[u<<1|1].sum;
}

void pushdown(int u){
	if(tr[u].tag){
		tr[u<<1].tag^=1;
		tr[u<<1].sum=tr[u<<1].r-tr[u<<1].l+1-tr[u<<1].sum;
		tr[u<<1|1].tag^=1;
		tr[u<<1|1].sum=tr[u<<1|1].r-tr[u<<1|1].l+1-tr[u<<1|1].sum;
		tr[u].tag=0;
	}
}

void build(int u,int l,int r){
	tr[u]={l,r,0,0};
	if(l==r) return;
	int m=(l+r)>>1;
	build(u<<1,l,m);
	build(u<<1|1,m+1,r);
}

void modify(int u,int l,int r){
	if(l<=tr[u].l&&tr[u].r<=r){
		tr[u].sum=tr[u].r-tr[u].l+1-tr[u].sum;
		tr[u].tag^=1;
		return;
	}
	pushdown(u);
	int m=(tr[u].l+tr[u].r)>>1;
	if(l<=m) modify(u<<1,l,r);
	if(r>m) modify(u<<1|1,l,r);
	pushup(u);
}

int query(int u,int l,int r){
	if(l<=tr[u].l&&tr[u].r<=r){
		return tr[u].sum;
	}
	pushdown(u);
	int m=(tr[u].l+tr[u].r)>>1;
	int res=0;
	if(l<=m) res+=query(u<<1,l,r);
	if(r>m) res+=query(u<<1|1,l,r);
	return res;
}

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;
	cin>>n;
	vector<vector<int>> e(n);
	vector<int> L(n),R(n);
	int dfn=0;
	for(int i=1;i<n;i++){
		int p;
		cin>>p;
		p--;
		e[p].push_back(i);
	}
	function<void(int)> dfs=[&](int x){
		L[x]=++dfn;
		for(auto y:e[x]){
			dfs(y);
		}
		R[x]=dfn;
	};
	dfs(0);
	build(1,1,n);
	for(int i=0;i<n;i++){
		int w;
		cin>>w;
		if(w) modify(1,L[i],L[i]);
	}
	int q;
	cin>>q;
	while(q--){
		string op;
		int x;
		cin>>op>>x;
		x--;
		if(op=="get"){
			cout<<query(1,L[x],R[x])<<'\n';
		}else{
			modify(1,L[x],R[x]);
		}
	}
	return 0;
}