树套树

Posted 2020-12-21 sduwh

树套树

一种思想，就是一棵树的节点是另一颗树。

在外面的叫外层树，在里面的叫内层树。

外层树一般是， 树状数组 ， 线段树

内层树一般是 平衡树 ， STL , 线段树

线段树套STL

/*
 * @Author: zhl
 * @Date: 2020-11-16 12:50:32
 */
#include<bits/stdc++.h>
#define lo (o<<1)
#define ro (o<<1|1)
#define mid (l+r>>1)
using namespace std;

const int N = 5e4 + 10, inf = 1e9;

multiset<int>s[N << 2];
int A[N];
void build(int o, int l, int r) {
	s[o].insert(inf); s[o].insert(-inf);
	for (int i = l; i <= r; i++) s[o].insert(A[i]);
	if (l == r)return;
	build(lo, l, mid);
	build(ro, mid + 1, r);
}
void updt(int o, int l, int r, int pos, int v) {
	s[o].erase(s[o].lower_bound(A[pos]));
	s[o].insert(v);
	if (l == r)return;
	if (pos <= mid) updt(lo, l, mid, pos, v);
	else updt(ro, mid + 1, r, pos, v);
}

int query(int o, int l, int r, int L, int R, int v) {
	if (L <= l and r <= R) return *prev(s[o].lower_bound(v));
	int ans = -inf;
	if (L <= mid)ans = max(ans, query(lo, l, mid, L, R, v));
	if (R > mid) ans = max(ans, query(ro, mid + 1, r, L, R, v));
	return ans;
}
int n, m;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)scanf("%d", A + i);
	build(1, 1, n);
	while (m--) {
		int op, a, b, x; scanf("%d", &op);
		if (op == 1) {
			scanf("%d%d", &a, &b);
			updt(1, 1, n, a, b);
			A[a] = b;
		}
		else {
			scanf("%d%d%d", &a, &b, &x);
			printf("%d
", query(1, 1, n, a, b, x));
		}
	}
}

线段树套平衡树

【树套树模板】二逼平衡树

很多棵树的时候可以开一个 root 数组就可以，这样可以不需要传引用，因为在splay的时候会更新 root数组

rotate 不可以任意顺序，会有影响

/*
 * @Author: zhl
 * @Date: 2020-11-16 13:51:18
 */

#include<bits/stdc++.h>
#define mid (l+r>>1)
#define lo (o<<1)
#define ro (o<<1|1)
using namespace std;

const int N = 2e6 + 10, inf = 0x7fffffff;

struct node {
	int s[2], size, p, v;
	void init(int _p, int _v) {
		p = _p; v = _v; size = 1;
	}
}tr[N];

int w[N], n, m, root[N], idx;

void push_up(int u) {
	tr[u].size = tr[tr[u].s[0]].size + tr[tr[u].s[1]].size + 1;
}
void rotate(int x) {
	int y = tr[x].p, z = tr[y].p;
	int k = tr[y].s[1] == x;
	tr[z].s[tr[z].s[1] == y] = x; tr[x].p = z;
	tr[y].s[k] = tr[x].s[k ^ 1]; tr[tr[x].s[k ^ 1]].p = y; //草这两行顺序不能换
	tr[x].s[k ^ 1] = y; tr[y].p = x;
	push_up(y), push_up(x);
}

void splay(int x, int k,int rt) {
	
	while (tr[x].p != k) {
		int y = tr[x].p, z = tr[y].p;
		if (z != k) {
			if ((tr[z].s[0] == y) ^ (tr[y].s[0] == x)) rotate(x);
			else rotate(y);
		}
		rotate(x);
	}
	if (!k)root[rt] = x;
}

void insert(int v, int rt) {
	int u = root[rt], p = 0;
	while (u) p = u, u = tr[u].s[v > tr[u].v];
	u = ++idx;
	if (p)tr[p].s[v > tr[p].v] = u;
	tr[u].init(p, v);
	splay(u, 0, rt);
}

int get_rank(int v, int rt) {
	int u = root[rt], res = 0;
	while (u) {
		if (v > tr[u].v) res += tr[tr[u].s[0]].size + 1, u = tr[u].s[1];
		else u = tr[u].s[0];
	}
	return res;
}
void build(int o, int l, int r) {
	insert(-inf, o); insert(inf, o);
	for (int i = l; i <= r; i++) {
		insert(w[i], o);
	}
	if (l == r)return;
	build(lo, l, mid);
	build(ro, mid + 1, r);
}

int query_rank(int o, int l, int r, int L, int R, int x) {
	if (L <= l and r <= R)return get_rank(x, o) - 1;
	int ans = 0;
	if (L <= mid)ans += query_rank(lo, l, mid, L, R, x);
	if (R > mid) ans += query_rank(ro, mid + 1, r, L, R, x);
	return ans;
}
void updt(int o, int l, int r, int pos, int v){
	int u = root[o];
	while (u) {
		if (tr[u].v == w[pos])break;
		if (w[pos] > tr[u].v)u = tr[u].s[1];
		if (w[pos] < tr[u].v) u = tr[u].s[0];
	}
	splay(u, 0, o);
	int ls = tr[u].s[0], rs = tr[u].s[1];
	while (tr[ls].s[1]) ls = tr[ls].s[1];
	while (tr[rs].s[0]) rs = tr[rs].s[0];
	splay(ls, 0, o); splay(rs, ls, o);
	tr[rs].s[0] = 0;
	push_up(rs); push_up(ls);
	insert(v, o);
	if (l == r)return; //不要忘记结束条件
	if (pos <= mid) {
		updt(lo, l, mid, pos, v);
	}
	else {
		updt(ro, mid + 1, r, pos, v);
	}
}
int get_pre(int x,int rt) {
	int u = root[rt], res = -inf;
	while (u) {
		if (tr[u].v >= x) u = tr[u].s[0];
		else res = tr[u].v, u = tr[u].s[1];
	}
	return res;
}
int get_suc(int x,int rt) {
	int u = root[rt], res = -inf;
	while (u) {
		if (tr[u].v <= x) u = tr[u].s[1];
		else res = tr[u].v, u = tr[u].s[0];
	}
	return res;
}

int query_pre(int o, int l, int r, int L, int R, int x) {
	if (L <= l and r <= R)return get_pre(x, o);
	int ans = -inf;
	if (L <= mid)ans = max(ans, query_pre(lo, l, mid, L, R, x));
	if (R > mid) ans = max(ans, query_pre(ro, mid + 1, r, L, R, x));
	return ans;
}

int query_suc(int o, int l, int r, int L, int R, int x) {
	if (L <= l and r <= R)return get_suc(x, o);
	int ans = inf;
	if (L <= mid)ans = min(ans, query_suc(lo, l, mid, L, R, x));
	if (R > mid) ans = min(ans, query_suc(ro, mid + 1, r, L, R, x));
	return ans;
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)scanf("%d", w + i);
	build(1, 1, n);
	while (m--) {
		int op, a, b, k, pos;
		scanf("%d", &op);
		if (op == 1) {
			scanf("%d%d%d", &a, &b, &k);
			printf("%d
", query_rank(1, 1, n, a, b, k) + 1);
		}
		else if (op == 2) {
			scanf("%d%d%d", &a, &b, &k);
			int l = 0, r = 1e8;
			while (l < r) {
				int m = l + r + 1 >> 1;
				if (query_rank(1, 1, n, a, b, m) + 1 <= k) {
					l = m;
				}
				else {
					r = m - 1;
				}
			}
			printf("%d
", r);
		}
		else if (op == 3) {
			scanf("%d%d", &pos, &k);
			updt(1, 1, n, pos, k);
			w[pos] = k;
		}
		else if (op == 4) {
			scanf("%d%d%d", &a, &b, &k);
			printf("%d
", query_pre(1, 1, n, a, b, k));
		}
		else if (op == 5) {
			scanf("%d%d%d", &a, &b, &k);
			printf("%d
", query_suc(1, 1, n, a, b, k));
		}
	}

}