P6812 「MCOI-02」Ancestor 先辈

思考满足条件的序列的本质。

如果对于序列 $a$ 的每个后缀 $b$，都有 $a$ 比 $b$ 屑，那么就是说，$a_1 \leq a_2, a_3, \cdots, a_n$，$a_2 \leq a_3, a_4, \cdots, a_n$……显然这是个非严格上升的序列。

所以线段树维护之，复杂度 $O(n \log n)$，注意 pushdown 和 merge 的写法即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

#define int long long

const int N = 1e6 + 5;

int a[N], n, m;

class SegmentTree
{
public:
	struct Node
	{
		int l, r, addtag, lval, rval;
		bool pd;
	}tr[N << 2];
	void pushup(int u)
	{
		tr[u].lval = tr[u << 1].lval, tr[u].rval = tr[u << 1 | 1].rval;
		tr[u].pd = (tr[u << 1].pd && tr[u << 1 | 1].pd && tr[u << 1].rval <= tr[u << 1 | 1].lval);
	}
	Node merge(Node a, Node b)
	{
		if (a.l == -1 && a.r == -1) return b;
		if (b.l == -1 && b.r == -1) return a;
		Node c;
		c.lval = a.lval;
		c.rval = b.rval;
		c.pd = (a.pd && b.pd && a.rval <= b.lval);
		c.l = a.l, c.r = b.r;
		return c;
	}
	void pushdown(int u)
	{
		if (tr[u].addtag)
		{
			tr[u << 1].addtag += tr[u].addtag;
			tr[u << 1].lval += tr[u].addtag;
			tr[u << 1].rval += tr[u].addtag;
			tr[u << 1 | 1].addtag += tr[u].addtag;
			tr[u << 1 | 1].lval += tr[u].addtag;
			tr[u << 1 | 1].rval += tr[u].addtag;
			tr[u].addtag = 0;
		}
	}
	void build(int u, int l, int r)
	{
		tr[u] = { l, r, 0, a[r], a[r], 1 };
		if (l == r) return;
		int mid = l + r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
		pushup(u);
	}
	void update(int u, int l, int r, int k)
	{
		if (tr[u].l >= l and tr[u].r <= r)
		{
			tr[u].addtag += k;
			tr[u].lval += k, tr[u].rval += k;
			return;
		}
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1;
		if (l <= mid) update(u << 1, l, r, k);
		if (r > mid) update(u << 1 | 1, l, r, k);
		pushup(u);
	}
	Node query(int u, int l, int r)
	{
		if (tr[u].l >= l and tr[u].r <= r)
		{
			return tr[u];
		}
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1;
		Node p;
		p = { -1, -1, 0, 0, 0, 1 };
		if (l <= mid) p = query(u << 1, l, r);
		if (r > mid)
		{
			p = merge(p, query(u << 1 | 1, l, r));
		}
		return p;
	}
}tr;

signed main()
{
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	tr.build(1, 1, n);
	while (m--)
	{
		int op, l, r;
		scanf("%lld%lld%lld", &op, &l, &r);
		r = min(r, n);
		if (op == 1)
		{
			int k;
			scanf("%lld", &k);
			tr.update(1, l, r, k);
		}
		else
		{
			if (tr.query(1, l, r).pd) printf("Yes\n");
			else printf("No\n");
		}
	}
	return 0;
}