模板 - 数据结构 - 无旋Treap / FHQ Treap
普通平衡树:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls(p) ch[p][0]
#define rs(p) ch[p][1]
const int MAXN = 100000 + 5;
int val[MAXN], ch[MAXN][2], rnd[MAXN], siz[MAXN], tot, root;
void Init() {
tot = root = 0;
}
void PushUp(int p) {
siz[p] = siz[ls(p)] + siz[rs(p)] + 1;
}
void SplitValue(int p, int v, int &x, int &y) {
if(!p) {
x = y = 0;
return;
}
if(v < val[p]) {
y = p;
SplitValue(ls(p), v, x, ls(p));
PushUp(y);
} else {
x = p;
SplitValue(rs(p), v, rs(p), y);
PushUp(x);
}
}
void SplitRank(int p, int rk, int &x, int &y) {
if(!p) {
x = y = 0;
return;
}
if(rk <= siz[ls(p)]) {
y = p;
SplitRank(ls(p), rk, x, ls(p));
PushUp(y);
} else {
x = p;
SplitRank(rs(p), rk - siz[ls(p)] - 1, rs(p), y);
PushUp(x);
}
}
int Merge(int x, int y) {
if(!x || !y)
return x | y;
if(rnd[x] < rnd[y]) {
rs(x) = Merge(rs(x), y);
PushUp(x);
return x;
} else {
ls(y) = Merge(x, ls(y));
PushUp(y);
return y;
}
}
int NewNode(int v) {
++tot;
ch[tot][0] = ch[tot][1] = 0;
val[tot] = v, rnd[tot] = rand();
siz[tot] = 1;
return tot;
}
void Insert(int &root, int v) {
int x = 0, y = 0;
SplitValue(root, v, x, y);
root = Merge(Merge(x, NewNode(v)), y);
}
void Remove(int &root, int v) {
int x = 0, y = 0, z = 0;
SplitValue(root, v, x, z);
SplitValue(x, v - 1, x, y);
y = Merge(ls(y), rs(y));
root = Merge(Merge(x, y), z);
}
int GetRank(int &root, int v) {
int x = 0, y = 0;
SplitValue(root, v - 1, x, y);
int rk = siz[x] + 1;
root = Merge(x, y);
return rk;
}
int GetValue(int &root, int rk) {
int x = 0, y = 0, z = 0;
SplitRank(root, rk, x, z);
SplitRank(x, rk - 1, x, y);
int v = val[y];
root = Merge(Merge(x, y), z);
return v;
}
int GetPrev(int &root, int v) {
int x = 0, y = 0;
SplitValue(root, v - 1, x, y);
int prev = GetValue(x, siz[x]);
root = Merge(x, y);
return prev;
}
int GetNext(int &root, int v) {
int x = 0, y = 0;
SplitValue(root, v, x, y);
int next = GetValue(y, 1);
root = Merge(x, y);
return next;
}
int main() {
int n;
scanf("%d", &n);
Init();
for(int i = 1; i <= n; ++i) {
int op, x;
scanf("%d%d", &op, &x);
switch(op) {
case 1:
Insert(root, x);
break;
case 2:
Remove(root, x);
break;
case 3:
printf("%d\n", GetRank(root, x));
break;
case 4:
printf("%d\n", GetValue(root, x));
break;
case 5:
printf("%d\n", GetPrev(root, x));
break;
case 6:
printf("%d\n", GetNext(root, x));
break;
}
}
return 0;
}
O(n)建树与回收:
//O(n)建树,返回新树的根
int st[MAXN], stop;
char buf[MAXN];
inline int Build(int n) {
stop = 0;
for(int i = 0; i < n; ++i) {
int tmp = NewNode(buf[i]), last = 0;
while(stop && rnd[st[stop]] > rnd[tmp]) {
last = st[stop];
PushUp(last);
st[stop--] = 0;
}
if(stop)
rs(st[stop]) = tmp;
ls(tmp) = last;
st[++stop] = tmp;
}
while(stop)
PushUp(st[stop--]);
return st[1];
}
//O(n)回收整棵树
inline void UnBuild(int p) {
if(!p)
return;
UnBuild(ls(p));
UnBuild(rs(p));
RecBin.push(p);
}
非递归查询:
int GetRank2(int p, int v) {
int rk = 1;
while(p) {
if(v < val[p])
p = ls(p);
else if(v == val[p])
p = ls(p);
else {
rk += siz[ls(p)] + 1;
p = rs(p);
}
}
return rk;
}
int GetValue2(int p, int rk) {
while(p) {
if(rk <= siz[ls(p)])
p = ls(p);
else if(rk == siz[ls(p)] + 1)
return val[p];
else {
rk -= siz[ls(p)] + 1;
p = rs(p);
}
}
}
int GetPrev2(int p, int v) {
int prev;
while(p) {
if(v <= val[p])
p = ls(p);
else {
prev = val[p];
p = rs(p);
}
}
return prev;
}
int GetNext2(int p, int v) {
int next;
while(p) {
if(v < val[p]) {
next = val[p];
p = ls(p);
} else
p = rs(p);
}
return next;
}
无旋Treap维护序列:
操作1翻转序列[L,R],操作2查询pos位置的字符。
#define ls(p) ch[p][0]
#define rs(p) ch[p][1]
const int MAXN = 1000000 + 5;
char val[MAXN];
int ch[MAXN][2], rnd[MAXN], siz[MAXN], tot, root;
bool rev[MAXN];
void Init() {
tot = root = 0;
}
void PushUp(int p) {
siz[p] = siz[ls(p)] + siz[rs(p)] + 1;
}
void PushDown(int p) {
if(rev[p]) {
swap(ls(p), rs(p));
rev[ls(p)] ^= 1;
rev[rs(p)] ^= 1;
rev[p] = 0;
}
}
void SplitRank(int p, int rk, int &x, int &y) {
if(!p) {
x = y = 0;
return;
}
PushDown(p);
if(rk <= siz[ls(p)]) {
y = p;
SplitRank(ls(p), rk, x, ls(p));
PushUp(y);
} else {
x = p;
SplitRank(rs(p), rk - siz[ls(p)] - 1, rs(p), y);
PushUp(x);
}
}
int Merge(int x, int y) {
if(!x || !y)
return x | y;
if(rnd[x] < rnd[y]) {
PushDown(x);
rs(x) = Merge(rs(x), y);
PushUp(x);
return x;
} else {
PushDown(y);
ls(y) = Merge(x, ls(y));
PushUp(y);
return y;
}
}
int NewNode(int v) {
++tot;
ch[tot][0] = ch[tot][1] = 0;
val[tot] = v, rnd[tot] = rand();
siz[tot] = 1;
return tot;
}
void ReverseLR(int l, int r) {
int x, y, z;
SplitRank(root, l - 1, x, y);
SplitRank(y, r + 1 - l, y, z);
PushDown(y);
rev[y] ^= 1;
PushUp(y);
root = Merge(Merge(x, y), z);
}
int ValuePos(int pos) {
int x, y, z;
SplitRank(root, pos - 1, x, z);
SplitRank(z, 1, y, z);
int ret = val[y];
root = Merge(x, Merge(y, z));
return ret;
}
需要搭配快读和线性建树才能卡过去。