Geiz-Revive > 洛谷-P5350-序列

洛谷-P5350-序列

2019-09-24

Splay

$\large \color{blue} Describe$

这题只有在数据随机的情况下4操作的复杂度才是正确的，因为数据随机，因此我们可以考虑平衡
树维护相同值域连续段，然后重点就是拆区间，我们考虑把在找节点时把一个连续段拆成3段x、y、z，把z作为y的后继，y作为x的后继，再返回y即可，其他思路都非常简单，主要考察代码能力。

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#define poi(x) (rs(fa(x)) == (x))
#define reverse(x) (swap(ls(x), rs(x)), t[x].filp ^= 1)
const int p = 1e9 + 7;
stack<int> st;
int rt, cnt, a[300005], len[300005];
struct node {
	int fa, child[2], tot, sum, w, size, len;
	bool filp, tag;
} t[3000005];
inl void upd(int x) {
	t[x].tot = ((t[ls(x)].tot + t[rs(x)].tot) % p + (1ll * t[x].w * t[x].len)) % p;
	t[x].size = t[ls(x)].size + t[rs(x)].size + t[x].len;
}
inl int newn(int w, int len) {
	re x;
	if (!st.empty())
		x = st.top(), st.pop();
	else
		x = ++cnt;
	t[x].size = t[x].tot = fa(x) = ls(x) = rs(x) = t[x].filp = t[x].sum = 0;
	t[x].w = w, t[x].len = len, upd(x);
	return x;
}
inl void rotate(int x) {
	re f = fa(x), gf = fa(f), fs = poi(x), gfs = poi(f), s = son(x, fs ^ 1);
	if (gf) son(gf, gfs) = x;
	son(f, fs) = s, son(x, fs ^ 1) = f;
	if (s) fa(s) = f;
	fa(f) = x, fa(x) = gf, upd(f);
}
inl void sum(int x, int w) { t[x].tot = (t[x].tot + 1ll * w * t[x].size) % p, t[x].w = (t[x].w + w) % p, t[x].sum = (t[x].sum + w) % p; }
inl void pushdown(int x) {
	if (t[x].sum) {
		if (ls(x)) sum(ls(x), t[x].sum);
		if (rs(x)) sum(rs(x), t[x].sum);
		t[x].sum = 0;
	}
	if (t[x].filp) {
		if (ls(x)) reverse(ls(x));
		if (rs(x)) reverse(rs(x));
		t[x].filp = 0;
	}
}
inl void push(int x) {
	if (fa(x)) push(fa(x));
	pushdown(x);
}
inl void splay(int x, int to = 0) {
	push(x);
	while (fa(x) != to) {
		if (fa(fa(x)) != to) poi(x) == poi(fa(x)) ? rotate(fa(x)) : rotate(x);
		rotate(x);
	}
	if (!to) rt = x;
	upd(x);
}
inl int split(int x, int k) {
	re y = newn(t[x].w, t[x].len - k);
	t[x].len = k;
	if (!rs(x))fa(rs(x) = y) = x;
	else {
		pushdown(x);
		re o = rs(x); pushdown(o);
		while (ls(o))pushdown(o = ls(o));
		fa(ls(o) = y) = o;
		while (o != x) upd(o), o = fa(o);
	}
	splay(y);
	return y;
}
inl void find(int k) {
	re x = rt;
	while (x) {
		pushdown(x);
		if (t[ls(x)].size >= k) x = ls(x);
		else {
			k -= t[ls(x)].size;
			if (t[x].len >= k) {
				if (k != t[x].len)split(x, k);
				if (k != 1)x = split(x, k - 1);
				return splay(x);
			}
			else k -= t[x].len, x = rs(x);
		}
	}
}
inl int nxt(int k, bool f) {
	find(k);
	re x = rt;
	x = son(x, f);
	while (son(x, f ^ 1))pushdown(x), x = son(x, f ^ 1);
	return x;
}
inl void build(int &k, int l, int r) {
	if (l > r) return;
	re mid = l + r >> 1;
	k = newn(a[mid], len[mid]);
	if (a[mid] == inf)t[k].tag = 1, t[k].w = 0, upd(k);
	build(ls(k), l, mid - 1), build(rs(k), mid + 1, r);
	if (ls(k))fa(ls(k)) = k; if (rs(k)) fa(rs(k)) = k;
	upd(k);
}
int num;
inl void travel(int x) {
	pushdown(x);
	if (ls(x)) travel(ls(x));
	a[++num] = t[x].w, len[num] = t[x].len;
	if (rs(x)) travel(rs(x));
}
inl void erase(int x) {
	if (ls(x)) erase(ls(x));
	st.push(x);
	if (rs(x)) erase(rs(x));
}
inl void change(int x, int w) {
	if (w == 1) {
		t[x].w = w, upd(x);
		return;
	}
	erase(x);
	re k = fa(x);
	ls(k) = fa(x) = 0;
	fa(ls(k) = newn(w, t[x].size)) = k;
	splay(ls(k));
}
inl void print(int x) {
	pushdown(x);
	if (ls(x))print(ls(x));
	if (!t[x].tag)for (re i = 1; i <= t[x].len; i++)writesp(t[x].w);
	if (rs(x))print(rs(x));
}
inl int cut(int a, int b) {
	re l, r;
	find(a), l = rt, find(b + 2), r = rt;
	splay(l), splay(r, l);
	return ls(r);
}
signed main() {
	re n = read<int>(), m = read<int>(), l, r, op, l1, r1, w, tmp1, tmp2, f;
	a[++num] = inf, len[num] = 1;
	for (re i = 1; i <= n; i++) a[++num] = read<int>() % p, len[num] = 1;
	a[++num] = inf, len[num] = 1;
	build(rt, 1, num);
	while (m--) {
		op = read<int>(), l = read<int>(), r = read<int>();
		switch (op) {
		case 1: {
			writeln(t[cut(l, r)].tot);
			break;
		}
		case 2: {
			w = read<int>() % p;
			change(cut(l, r), w);
			break;
		}
		case 3: {
			w = read<int>() % p;
			sum(cut(l, r), w);
			break;
		}
		case 4: {
			l1 = read<int>(), r1 = read<int>();
			num = 0, travel(cut(l, r));
			build(tmp1, 1, num);
			erase(tmp2 = cut(l1, r1));
			ls(fa(tmp2)) = 0, f = fa(tmp2), fa(tmp2) = 0;
			fa(ls(f) = tmp1) = f;
			splay(ls(f));
			break;
		}
		case 5: {
			l1 = read<int>(), r1 = read<int>();
			num = 0, travel(cut(l, r));
			build(tmp1, 1, num);
			tmp2 = cut(l1, r1);
			ls(fa(tmp2)) = 0, f = fa(tmp2), fa(tmp2) = 0;
			fa(ls(f) = tmp1) = f;
			splay(ls(f));
			f = cut(l, r);
			ls(fa(f)) = tmp2, fa(tmp2) = fa(f), fa(f) = 0;
			erase(f);
			break;
		}
		case 6: {
			f = cut(l, r);
			reverse(f);
			break;
		}
		}
	}
	print(rt);
}