FHQ-treap，即无旋 treap，又称分裂合并 treap，支持维护序列，可持久化等特性。
FHQ-treap 有两个核心操作，
vector<int> rub;
int newnod(int x) {
int u;
if (!rub.empty()) {
u = rub.back();
rub.pop_back();
}
else {
u = ++ tot;
}
t[u].siz = 1;
t[u].ls = t[u].rs = 0;
t[u].val = x;
t[u].pai = rand();
return u;
}
分裂
分裂有两种，一种是按照节点个数来分裂，另一种是按照权值大小来分裂。
一般最常用的是按照节点个数分裂，但是按照权值大小分裂也会用到。
一般进行操作，我们的通用方法是将被操作点单独分裂成一棵树，对这棵树进行操作。
按照节点数量分裂的代码。
void split_rk(int u, int k, int &x, int &y) {
if (u == 0) {
x = y = 0;
return ;
}
if (t[lc].siz + 1 <= k) {
x = u;
split_rk(rc, k - t[lc].siz - 1, t[u].rs, y);
}
else {
y = u;
split_rk(lc, k, x, t[u].ls);
}
pushup(u);
}
按照权值分裂的代码。
void split_val(int u, int v, int &x, int &y) { // x 和 y 是传参类型
if (u == 0) {
x = y = 0;
return ;
}
if (t[u].val <= v) {
x = u;
split_val(rc, v, rc, y);
}
else {
y = u;
split_val(lc, v, x, lc);
}
pushup(u);
}
合并
在旋转 treap 中，我们借助旋转操作来维护堆的性质，同时旋转时还不能改变树的性质。在无旋 treap 中，我们用合并达到相同的效果。
因为两个 treap 已经有序，所以我们在合并的时候只需要考虑把哪个树「放在上面」，把哪个「放在下面」，也就是是需要判断将哪个一个树作为子树。显然，根据堆的性质，我们需要把 \(pai\) 小的放在上面（这里采用小根堆）。
同时，我们还需要满足搜索树的性质。设 \(u < v\)，若 \(u\) 的 \(pai\) 小于 \(v\) 的，那么 \(u\) 即为新根结点，并且 \(v\) 因为值比 \(u\) 更大，应与 \(u\) 的右子树合并；反之，则 \(v\) 作为新根结点，然后因为 \(u\) 的值比 \(v\) 小，与 \(v\) 的左子树合并。
int Merge(int x, int y) {
if (!x || !y) {
return x + y;
}
if (t[x].pai < t[y].pai) {
t[x].rs = Merge(t[x].rs, y);
pushup(x);
return x;
}
else {
t[y].ls = Merge(x, t[y].ls);
pushup(y);
return y;
}
}
基本操作
插入
将新节点要插入的位置分裂出来，然后合并即可。
void Insert(int x) {
int u = newnod(x);
int t1, t2;
split_val(rt, x, t1, t2);
rt = Merge(Merge(t1, u), t2);
}
删除
将要删除的节点分裂出来，将两边的子树合并即可。
void Erase(int x) {
int t1, t2, t3;
split_val(rt, x - 1, t1, t2);
split_val(t2, x, t2, t3);
rub.emplace_back(t2);
t2 = Merge(t[t2].ls, t[t2].rs);
rt = Merge(Merge(t1, t2), t3);
}
查找排名
将要查找的点按照权值分裂出来，前面分裂出去的树的大小 \(+ 1\) 就是排名。
int getrank(int x) {
int t1, t2, rk;
split_val(rt, x - 1, t1, t2);
rk = t[t1].siz + 1;
rt = Merge(t1, t2);
ans ^= rk;
las = rk;
return rk;
}
查找排名为 \(x\) 的节点的权值
将要查找的点按照节点个数分裂出来，进行操作。
int getval(int x) {
int t1, t2, t3, val;
split_rk(rt, x - 1, t1, t2);
split_rk(t2, 1, t2, t3);
val = t[t2].val;
ans ^= val;
las = val;
Merge(Merge(t1, t2), t3);
return val;
}
查找前驱后继
利用分裂来查找。
int pre(int x) {
int t1, t2, t3, pre;
split_val(rt, x - 1, t1, t2);
split_rk(t1, t[t1].siz - 1, t1, t3);
pre = t[t3].val;
rt = Merge(Merge(t1, t3), t2);
ans ^= pre;
las = pre;
return pre;
}
int nxt(int x) {
int t1, t2, t3, nxt;
split_val(rt, x, t1, t2);
split_rk(t2, 1, t2, t3);
nxt = t[t2].val;
rt = Merge(Merge(t1, t2), t3);
ans ^= nxt;
las = nxt;
return nxt;
}
区间操作
P3391 【模板】文艺平衡树 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
区间翻转练习题
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lc t[u].ls
#define rc t[u].rs
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 1e5 + 5;
int n, m, rt, tot;
vector<int> rub;
struct node {
int val, tag;
int ls, rs, siz, pai;
} t[N << 1];
inline void pushup(int u) {
t[u].siz = t[lc].siz + t[rc].siz + 1;
}
inline void pushdown(int u) {
if (!t[u].tag) {
return ;
}
if (lc) t[lc].tag ^= 1;
if (rc) t[rc].tag ^= 1;
swap(t[u].ls, t[u].rs);
t[u].tag = 0;
}
int newnod(int x) {
int u = ++ tot;
t[u].siz = 1;
t[u].ls = t[u].rs = t[u].tag = 0;
t[u].val = x;
t[u].pai = rand();
return u;
}
void split_rk(int u, int k, int &x, int &y) {
if (!u) {
x = y = 0;
return ;
}
pushdown(u);
if (t[lc].siz + 1 <= k) {
x = u;
split_rk(rc, k - t[lc].siz - 1, rc, y);
}
else {
y = u;
split_rk(lc, k, x, lc);
}
pushup(u);
}
int Merge(int x, int y) {
if (!x || !y) {
return x + y;
}
if (t[x].pai < t[y].pai) {
pushdown(x);
t[x].rs = Merge(t[x].rs, y);
pushup(x);
return x;
}
else {
pushdown(y);
t[y].ls = Merge(x, t[y].ls);
pushup(y);
return y;
}
}
void print(int u) {
if (!u) return ;
pushdown(u);
print(t[u].ls);
printf("%d ", t[u].val);
print(t[u].rs);
}
int main() {
srand(time(NULL));
n = read<int>(), m = read<int>();
for (int i = 1; i <= n; ++ i) {
rt = Merge(rt, newnod(i));
}
for (int i = 1, l, r; i <= m; ++ i) {
l = read<int>(), r = read<int>();
int t1, t2, t3;
split_rk(rt, l - 1, t1, t2);
split_rk(t2, r - l + 1, t2, t3);
t[t2].tag ^= 1;
rt = Merge(t1, Merge(t2, t3));
}
print(rt);
return 0;
}
P2042 [NOI2005] 维护数列 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
区间操作的练习好题，涉及线段树操作。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lc (t[u].ls)
#define rc (t[u].rs)
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 1e6 + 6;
mt19937 rnd(time(0));
int n, m, tot, rt;
int a[N];
vector<int> rub;
struct node {
int pai, ls, rs, siz;
ll val, sum, mx, maxpre, maxlas, tag;
bool tag1, tag2;
} t[N];
int New(int x) {
int u;
if (!rub.empty()) {
u = rub.back();
rub.pop_back();
} else {
u = ++ tot;
}
t[u].sum = t[u].val = (t[u].mx = x);
t[u].maxpre = t[u].maxlas = max(0, x);
t[u].siz = 1;
t[u].pai = rnd();
t[u].tag1 = t[u].tag2 = (t[u].tag = 0);
t[u].ls = t[u].rs = 0;
return u;
}
void pushup(int u) {
if (!u) return;
t[u].siz = t[lc].siz + t[rc].siz + 1;
t[u].sum = t[lc].sum + t[rc].sum + t[u].val;
t[u].maxpre = max(max(t[lc].maxpre, t[lc].sum + t[u].val + t[rc].maxpre), 0ll);
t[u].maxlas = max(max(t[rc].maxlas, t[rc].sum + t[u].val + t[lc].maxlas), 0ll);
t[u].mx = max(0ll, t[lc].maxlas + t[rc].maxpre) + t[u].val;
if (lc) t[u].mx = max(t[u].mx, t[lc].mx);
if (rc) t[u].mx = max(t[u].mx, t[rc].mx);
}
void cover(int u, ll c) {
t[u].val = t[u].tag = c;
t[u].sum = t[u].siz * c;
t[u].maxpre = t[u].maxlas = max(0ll, t[u].sum);
t[u].mx = max(c, t[u].sum);
t[u].tag1 = 1;
}
void Reverse(int u) {
if (!u) return ;
swap(lc, rc);
swap(t[u].maxpre, t[u].maxlas);
t[u].tag2 ^= 1;
}
void pushdown(int u) {
if (!u) return ;
if (t[u].tag2) {
if (lc) {
Reverse(lc);
}
if (rc) {
Reverse(rc);
}
t[u].tag2 = 0;
}
if (t[u].tag1) {
if (lc) {
cover(lc, t[u].tag);
}
if (rc) {
cover(rc, t[u].tag);
}
t[u].tag = t[u].tag1 = 0;
}
}
void split(int u, int k, int &x, int &y) {
if (!u) {
x = y = 0;
return;
}
pushdown(u);
if (t[lc].siz < k) {
x = u;
split(rc, k - t[lc].siz - 1, rc, y);
} else {
y = u;
split(lc, k, x, lc);
}
pushup(u);
}
int Merge(int x, int y) {
if (!x || !y) {
return x + y;
}
if (t[x].pai < t[y].pai) {
pushdown(x);
t[x].rs = Merge(t[x].rs, y);
pushup(x);
return x;
} else {
pushdown(y);
t[y].ls = Merge(x, t[y].ls);
pushup(y);
return y;
}
}
int add(int l, int r) {
if (l != r) {
int mid = (l + r) >> 1;
return Merge(add(l, mid), add(mid + 1, r));
}
return New(a[l]);
}
void Erase(int u) {
if (!u) return ;
rub.emplace_back(u);
if (lc) {
Erase(lc);
}
if (rc) {
Erase(rc);
}
}
void print(int u) {
if (!u) return ;
pushdown(u);
print(lc);
print(rc);
}
int main() {
n = read<int>(), m = read<int>();
for (int i = 1; i <= n; ++ i) {
a[i] = read<int>();
}
rt = Merge(rt, add(1, n));
string op;
for (int i = 1, t1, t2, t3; i <= m; ++ i) {
cin >> op;
if (op == "INSERT") {
int pos = read<int>(), len = read<int>();
split(rt, pos, t1, t2);
for (int i = 1; i <= len; ++ i) {
a[i] = read<int>();
}
rt = Merge(Merge(t1, add(1, len)), t2);
}
if (op == "DELETE") {
int pos = read<int>(), len = read<int>();
split(rt, pos - 1, t1, t2);
split(t2, len, t2, t3);
Erase(t2);
rt = Merge(t1, t3);
}
if (op == "MAKE-SAME") {
int pos = read<int>(), len = read<int>(), v = read<int>();
split(rt, pos - 1, t1, t2);
split(t2, len, t2, t3);
cover(t2, v);
rt = Merge(Merge(t1, t2), t3);
}
if (op == "REVERSE") {
int pos = read<int>(), len = read<int>();
split(rt, pos - 1, t1, t2);
split(t2, len, t2, t3);
Reverse(t2);
rt = Merge(Merge(t1, t2), t3);
}
if (op == "GET-SUM") {
int pos = read<int>(), len = read<int>();
split(rt, pos - 1, t1, t2);
split(t2, len, t2, t3);
printf("%lld\n", t[t2].sum);
rt = Merge(Merge(t1, t2), t3);
}
if (op == "MAX-SUM") {
printf("%lld\n", t[rt].mx);
}
}
return 0;
}
P2596 [ZJOI2006] 书架 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
模板题
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lc t[u].ls
#define rc t[u].rs
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 8e4 + 5;
mt19937 rnd(time(0));
int n, m, tot, rt;
int Id[N];
struct node {
int val, siz, pai;
int ls, rs, fa;
} t[N << 1];
void pushup(int u) {
t[u].siz = t[lc].siz + t[rc].siz + 1;
t[lc].fa = t[rc].fa = u;
}
int New(int x) {
int u = ++ tot;
t[u].siz = 1;
t[u].val = x;
t[u].pai = rnd();
t[u].ls = t[u].rs = t[u].fa = 0;
Id[x] = u;
return u;
}
int Find(int u) {
int res = t[t[u].ls].siz + 1;
for (; u != rt; u = t[u].fa) {
if (t[t[u].fa].rs == u) {
res += t[t[t[u].fa].ls].siz + 1;
}
}
return res;
}
void split_rk(int u, int k, int &x, int &y) {
if (!u) {
x = y = 0;
return ;
}
if (t[lc].siz < k) {
x = u;
split_rk(rc, k - t[lc].siz - 1, rc, y);
} else {
y = u;
split_rk(lc, k, x, lc);
}
pushup(u);
}
int Merge(int x, int y) {
if (!x || !y) {
return x + y;
}
if (t[x].pai < t[y].pai) {
t[x].rs = Merge(t[x].rs, y);
pushup(x);
return x;
} else {
t[y].ls = Merge(x, t[y].ls);
pushup(y);
return y;
}
}
int main() {
n = read<int>(), m = read<int>();
for (int i = 1; i <= n; ++ i) {
rt = Merge(rt, New(read<int>()));
}
for (int i = 1, x, t1, t2, t3, t4; i <= m; ++ i) {
string op;
cin >> op;
x = read<int>();
if (op == "Top") {
int k = Find(Id[x]);
split_rk(rt, k - 1, t1, t2);
split_rk(t2, 1, t2, t3);
rt = Merge(Merge(t2, t1), t3);
}
if (op == "Bottom") {
int k = Find(Id[x]);
split_rk(rt, k - 1, t1, t2);
split_rk(t2, 1, t2, t3);
rt = Merge(Merge(t1, t3), t2);
}
if (op == "Insert") {
int y = read<int>();
int k = Find(Id[x]);
if (y > 0) {
split_rk(rt, k - 1, t1, t2);
split_rk(t2, 1, t2, t3);
split_rk(t3, y, t3, t4);
rt = Merge(Merge(t1, t3), Merge(t2, t4));
} else {
split_rk(rt, k - 1, t1, t2);
split_rk(t2, 1, t2, t3);
split_rk(t1, k + y - 1, t1, t4);
rt = Merge(Merge(t1, t2), Merge(t4, t3));
}
}
if (op == "Ask") {
cout << Find(Id[x]) - 1 << '\n';
}
if (op == "Query") {
split_rk(rt, x - 1, t1, t2);
split_rk(t2, 1, t2, t3);
cout << t[t2].val << '\n';
rt = Merge(Merge(t1, t2), t3);
}
}
return 0;
}
P3369 【模板】普通平衡树 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define lc t[u].ls
#define rc t[u].rs
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 2e5 + 5;
int n, tot, top, rt;
vector<int> rub;
struct node {
int val, pai, siz;
int ls, rs;
} t[N];
inline int newnod(int x) {
int u;
if (!rub.empty()) {
u = rub.back();
rub.pop_back();
}
else {
u = ++ tot;
}
t[u].siz = 1;
t[u].ls = t[u].rs = 0;
t[u].val = x;
t[u].pai = rand();
return u;
}
inline void pushup(int u) {
t[u].siz = t[lc].siz + 1 + t[rc].siz;
}
void split_rk(int u, int k, int &x, int &y) {
if (u == 0) {
x = y = 0;
return ;
}
if (t[lc].siz + 1 <= k) {
x = u;
split_rk(rc, k - t[lc].siz - 1, t[u].rs, y);
}
else {
y = u;
split_rk(lc, k, x, t[u].ls);
}
pushup(u);
}
void split_val(int u, int v, int &x, int &y) {
if (u == 0) {
x = y = 0;
return ;
}
if (t[u].val <= v) {
x = u;
split_val(rc, v, rc, y);
}
else {
y = u;
split_val(lc, v, x, lc);
}
pushup(u);
}
int Merge(int x, int y) {
if (!x || !y) {
return x + y;
}
if (t[x].pai < t[y].pai) {
t[x].rs = Merge(t[x].rs, y);
pushup(x);
return x;
}
else {
t[y].ls = Merge(x, t[y].ls);
pushup(y);
return y;
}
}
inline void Insert(int x) {
int u = newnod(x);
int t1, t2;
split_val(rt, x, t1, t2);
rt = Merge(Merge(t1, u), t2);
}
inline void Erase(int x) {
int t1, t2, t3;
split_val(rt, x - 1, t1, t2);
split_val(t2, x, t2, t3);
rub.emplace_back(t2);
t2 = Merge(t[t2].ls, t[t2].rs);
rt = Merge(Merge(t1, t2), t3);
}
inline int getrank(int x) {
int t1, t2, rk;
split_val(rt, x - 1, t1, t2);
rk = t[t1].siz + 1;
rt = Merge(t1, t2);
return rk;
}
inline int getval(int x) {
int t1, t2, t3, val;
split_rk(rt, x - 1, t1, t2);
split_rk(t2, 1, t2, t3);
val = t[t2].val;
Merge(Merge(t1, t2), t3);
return val;
}
inline int pre(int x) {
int t1, t2, t3, pre;
split_val(rt, x - 1, t1, t2);
split_rk(t1, t[t1].siz - 1, t1, t3);
pre = t[t3].val;
rt = Merge(Merge(t1, t3), t2);
return pre;
}
inline int las(int x) {
int t1, t2, t3, las;
split_val(rt, x, t1, t2);
split_rk(t2, 1, t2, t3);
las = t[t2].val;
rt = Merge(Merge(t1, t2), t3);
return las;
}
int main() {
n = read<int>();
for (int i = 1, x, op; i <= n; ++ i) {
op = read<int>(), x = read<int>();
switch(op) {
case 1 :
Insert(x);
break ;
case 2 :
Erase(x);
break ;
case 3 :
printf("%d\n", getrank(x));
break ;
case 4 :
printf("%d\n", getval(x));
break ;
case 5 :
printf("%d\n", pre(x));
break ;
case 6 :
printf("%d\n", las(x));
break ;
}
}
return 0;
}