FHQ-Treap小记

写在前面

参考资料&&orz gyh

嘤

博主是在$gyh$大佬的博客的帮助下，学会了$\text{FHQ-Treap}$咋么写的，博文链接如下

Luckyblock—FHQ-Treap

感谢$gyh$大佬，$gyh$永远滴神！

另外的参考资料

Shiina_Mashiro—平衡树（Splay、fhq Treap）

ctjcalc—【数据结构】FHQ Treap 详解

关于

$\text{FHQ-Treap}$，又名无旋$\text{Treap}$，是一种不需要旋转的平衡树~~（废话！）~~，是由范浩强基于$\text{Treap}$发明的，以分裂、合并为核心操作，有码量小、易理解的优势，其操作方式使它天生支持维护序列、可持久化等特性。

直接上操作和代码吧，暂时还没有写操作的原理，以后会补

操作

---

核心操作

分裂

分裂分为按值分裂和按大小分裂两种，下面给出的代码是按值分裂

void split(int rt, int val, int &x, int &y) {
	if (!rt) { x = y = 0; return; }
	if (t[rt].val <= val) x = rt, split(t[rt].son[1], val, t[rt].son[1], y);
	else y = rt, split(t[rt].son[0], val, x, t[rt].son[0]);
	update(rt);
}

合并

int merge(int x, int y) {
	if (!x || !y) return x + y;
	if (t[x].ran < t[y].ran) {
		t[x].son[1] = merge(t[x].son[1], y), update(x);
		return x;
	}
	else {
		t[y].son[0] = merge(x, t[y].son[0]), update(y);
		return y;
	}
}

---

其他操作

新建节点及更新

void update(int rt) {
	t[rt].siz = t[t[rt].son[0]].siz + t[t[rt].son[1]].siz + 1;
}

int new_node(int val) {
	t[++tot].val = val, t[tot].siz = 1, t[tot].ran = rand();
	return tot;
}

插入节点

void insert(int val) {
	split(root, val, X, Y);
	root = merge(merge(X, new_node(val)), Y);
}

删除节点

void delte(int val) {
	split(root, val, X, Z), split(X, val - 1, X, Y);
	Y = merge(t[Y].son[0], t[Y].son[1]);
	root = merge(merge(X, Y), Z);
}

查询值为val的数的排名

void rank(int val) {
	split(root, val - 1, X, Y);
	cout << t[X].siz + 1 << '\n';
	root = merge(X, Y);
}

查询第k大值

int kth(int rt, int key) {
	while (1) {
		if (key <= t[t[rt].son[0]].siz) { rt = t[rt].son[0]; continue; }
		if (key == t[t[rt].son[0]].siz + 1) return rt;
		key -= t[t[rt].son[0]].siz + 1;
		rt = t[rt].son[1];
	}
}

查询前驱pre

void pre(int val) {
	split(root, val - 1, X, Y);
	cout << t[kth(X,  t[X].siz)].val << '\n';
	root = merge(X, Y);
}

查询后继suc

void suc(int val) {
	split(root, val, X, Y);
	cout << t[kth(Y, 1)].val << '\n';
	root = merge(X, Y);
}

代码

本代码为洛谷 P3369 【模板】普通平衡树的代码

/*
Author:loceaner
*/
#include 
#include 
#include 
#include 
#include 
using namespace std;

const int A = 5e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

int n, tot, root, X, Y, Z;
struct node { int son[2], siz, ran, val; } t[A]; 

void update(int rt) {
	t[rt].siz = t[t[rt].son[0]].siz + t[t[rt].son[1]].siz + 1;
}

int new_node(int val) {
	t[++tot].val = val, t[tot].siz = 1, t[tot].ran = rand();
	return tot;
}

void split(int rt, int val, int &x, int &y) {
	if (!rt) { x = y = 0; return; }
	if (t[rt].val <= val) x = rt, split(t[rt].son[1], val, t[rt].son[1], y);
	else y = rt, split(t[rt].son[0], val, x, t[rt].son[0]);
	update(rt);
}

int merge(int x, int y) {
	if (!x || !y) return x + y;
	if (t[x].ran < t[y].ran) {
		t[x].son[1] = merge(t[x].son[1], y), update(x);
		return x;
	}
	else {
		t[y].son[0] = merge(x, t[y].son[0]), update(y);
		return y;
	}
}

void insert(int val) {
	split(root, val, X, Y);
	root = merge(merge(X, new_node(val)), Y);
}

void delte(int val) {
	split(root, val, X, Z), split(X, val - 1, X, Y);
	Y = merge(t[Y].son[0], t[Y].son[1]);
	root = merge(merge(X, Y), Z);
}

void rank(int val) {
	split(root, val - 1, X, Y);
	cout << t[X].siz + 1 << '\n';
	root = merge(X, Y);
}

int kth(int rt, int key) {
	while (1) {
		if (key <= t[t[rt].son[0]].siz) { rt = t[rt].son[0]; continue; }
		if (key == t[t[rt].son[0]].siz + 1) return rt;
		key -= t[t[rt].son[0]].siz + 1;
		rt = t[rt].son[1];
	}
}

void pre(int val) {
	split(root, val - 1, X, Y);
	cout << t[kth(X,  t[X].siz)].val << '\n';
	root = merge(X, Y);
}

void suc(int val) {
	split(root, val, X, Y);
	cout << t[kth(Y, 1)].val << '\n';
	root = merge(X, Y);
}

int main() {
	n = read();
	for (int i = 1, opt, x; i <= n; i++) {
		opt = read(), x = read();
		if (opt == 1) insert(x);
		else if (opt == 2) delte(x);
		else if (opt == 3) rank(x);
		else if (opt == 4) cout << t[kth(root, x)].val << '\n';
		else if (opt == 5) pre(x);
		else if (opt == 6) suc(x);
	}
	return 0;
}