洛谷 P3919 【模板】可持久化线段树 1（可持久化数组）

主席树板子
刚学完可持久化，趁热打铁

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

//#define int long long
#define fr first
#define se second

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

const int INF = 0X3f3f3f3f, N = 1e6 + 10, MOD = 1e9 + 10;

int a[N];

int root[N], idx;

struct NODE {
	int l, r;
	int val;
} tr[4 * N + 4 * 20 * N];

int build(int l, int r) {
	int q = ++idx;
	if (l == r) {
		tr[q].val = a[l];
		return q;
	} else {
		int mid = l + r >> 1;
		tr[q].l = build(l, mid);
		tr[q].r = build(mid + 1, r);
		return q;
	}
}

int insert(int p, int l, int r, int pos, int x) {
	int q = ++idx;
	tr[q] = tr[p];
	if (l == r) {
		if (x != INF) tr[q].val = x;
		return q;
	} else {
		int mid = l + r >> 1;
		if (pos <= mid) tr[q].l = insert(tr[p].l, l, mid, pos, x);
		else tr[q].r = insert(tr[p].r, mid + 1, r, pos, x);
		return q;
	}
}

int query(int q, int l, int r, int pos) {
	if (l == r) return tr[q].val;
	else {
		int mid = l + r >> 1;
		if (pos <= mid) return query(tr[q].l, l, mid, pos);
		else return query(tr[q].r, mid + 1, r, pos);
	}
}

void work() {
	int n, m;
	cin >> n >> m;

	for (int i = 1; i <= n; i++) cin >> a[i];

	root[0] = build(1, n);

	int v, op, pos, x;
	for (int i = 1; i <= m; i++) {
		scanf("%d%d%d", &v, &op, &pos);
		if (op == 1) {
			scanf("%d", &x);
			root[i] = insert(root[v], 1, n, pos, x);
		} else {
			printf("%d\n", query(root[v], 1, n, pos));
			root[i] = insert(root[v], 1, n, pos, INF);
		}
	}
}

int main() {

	work();

	return 0;
}