想试试指针的写法

struct node {
  	node *ch[2];
  	int key;

  	node(){
  		key = 0;
  		ch[0] = ch[1] = nullptr;
  	}
  	
  	node(node *_node){
    	key = _node -> key;
    	ch[0] = _node -> ch[0], ch[1] = _node -> ch[1];
	}
};

然后就在#2 MLE了 可能因为本人比较菜，也没想到进一步优化的办法 代码在下方，请大佬指教

// Problem: P3919 【模板】可持久化线段树 1（可持久化数组）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3919
// Memory Limit: 500 MB
// Time Limit: 1500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// created on Laptop of Lucian Xu

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <utility>
#include <unordered_map>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<int, LL> PIL;

#define rep(i, l, r) for(auto i = (l); i <= (r); i++)
#define per(i, r, l) for(auto i = (r); i >= (l); i--)
#define ff first
#define ss second
#define m_p make_pair
#define p_b push_back
#define endl '\n'
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

const int N = 1e6+10;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;

int n, m, k, x, vi, op, a[N];

struct node {
  	node *ch[2];
  	int key;

  	node(){
  		key = 0;
  		ch[0] = ch[1] = nullptr;
  	}
  	
  	node(node *_node){
    	key = _node -> key;
    	ch[0] = _node -> ch[0], ch[1] = _node -> ch[1];
	}
};

struct segment_tree{
	node *root[N];
	
	node *build(int l, int r){
		node *new_node; new_node = new node();
		if(l == r){
			new_node -> key = a[l];
			return new_node;
		}
		int mid = (l + r) >> 1;
		new_node -> ch[0] = build(l, mid);
		new_node -> ch[1] = build(mid + 1, r);
		return new_node;
	}
	
	// a[k] 改成 x //
	node *modify(node *p, int l, int r, int k, int x){
		node *new_node;
		new_node = new node(p);
		if(l == r){
			new_node -> key = x;
			return new_node;
		}
		int mid = (l + r) >> 1;
		if(k <= mid) new_node -> ch[0] = modify(new_node -> ch[0], l, mid, k, x);
		else new_node -> ch[1] = modify(new_node -> ch[1], mid + 1, r, k, x);
		return new_node;
	}
	
	// 询问 p 为根节点的版本的 a[k] //
	int query(node *p, int l, int r, int k){
		if(l == r){
			return p -> key;
		}
		int mid = (l + r) >> 1;
		if(k <= mid) return query(p -> ch[0], l, mid, k);
		else return query(p -> ch[1], mid + 1, r, k);
	}
};

segment_tree tr;

int main(){
	
	ios::sync_with_stdio(false);
 	cin.tie(0);
 	cout.tie(0);
	
    cin >> n >> m;
    rep(i, 1, n) cin >> a[i];
    tr.root[0] = tr.build(1, n);
    rep(i, 1, m){
    	cin >> vi >> op;
    	if(op == 1){
    		cin >> k >> x;
    		tr.root[i] = tr.modify(tr.root[vi], 1, n, k, x);
    	}
    	else{
    		cin >> k;
    		tr.root[i] = tr.root[vi];
    		cout << tr.query(tr.root[vi], 1, n, k) << endl;
    	}
    }
    return 0;
}