Problem
查询区间第k大，但保证区间不互相包含（可以相交）
Solution
只需要对每个区间左端点进行排序，那它们的右端点必定单调递增，不然会出现区间包含的情况。
所以我们暴力对下一个区间加上这个区间没有的点，删去下个区间没有的点。
因为每个点最多被加入，删除1次，所以时间复杂度为O（nlogn）
Notice
当相邻两段区间不相交时，那么我们要先加入点，在删去点。
Code
非旋转Treap
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100000, M = 50000;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int point = 0, T[N + 5], root, ans[M + 5];
struct Node
{
int left, right, ask, id;
}Q[M + 5];
struct node
{
int Size[N + 5], Val[N + 5], Level[N + 5], Son[2][N + 5];
inline void up(int u)
{
Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + 1;
}
int Newnode(int v)
{
int u = ++point;
Val[u] = v, Level[u] = rand();
Size[u] = 1, Son[0][u] = Son[1][u] = 0;
return u;
}
int Merge(int X, int Y)
{
if (X * Y == 0) return X + Y;
if (Level[X] < Level[Y])
{
Son[1][X] = Merge(Son[1][X], Y);
up(X); return X;
}
else
{
Son[0][Y] = Merge(X, Son[0][Y]);
up(Y); return Y;
}
}
void Split(int u, int t, int &x, int &y)
{
if (!u)
{
x = y = 0;
return;
}
if (Val[u] <= t) x = u, Split(Son[1][u], t, Son[1][u], y);
else y = u, Split(Son[0][u], t, x, Son[0][u]);
up(u);
}
void Build(int l, int r)
{
int last, s[N + 5], top = 0;
rep(i, l, r)
{
int u = Newnode(T[i]);
last = 0;
while (top && Level[s[top]] > Level[u])
{
up(s[top]);
last = s[top--];
}
if (top) Son[1][s[top]] = u;
Son[0][u] = last;
s[++top] = u;
}
while (top) up(s[top--]);
root = s[1];
}
int Find_num(int u, int t)
{
if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
else if (t <= Size[Son[0][u]] + 1) return u;
else return Find_num(Son[1][u], t - Size[Son[0][u]] - 1);
}
void Insert(int v)
{
int t = Newnode(v), x, y;
Split(root, v, x, y);
root = Merge(Merge(x, t), y);
}
void Delete(int v)
{
int x, y, z;
Split(root, v, x, z), Split(x, v - 1, x, y);
root = Merge(Merge(x, Merge(Son[0][y], Son[1][y])), z);
}
}Treap;
int cmp(Node X, Node Y)
{
return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
}
int sqz()
{
int n = read(), m = read();
rep(i, 1, n) T[i] = read();
rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
sort(Q + 1, Q + m + 1, cmp);
rep(i, Q[1].left, Q[1].right) Treap.Insert(T[i]);
ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
rep(i, 2, m)
{
rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(T[j]);
rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(T[j]);
ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
}
rep(i, 1, m) printf("%d\n", ans[i]);
return 0;
}
旋转Treap
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100000, M = 50000;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int point = 0, T[N + 5], root, ans[M + 5];
struct node
{
int left, right, ask, id;
}Q[M + 5];
int cmp(node X, node Y)
{
return X.left < Y.left || (X.left == Y.left && X.right < Y.right);
}
struct Node
{
int Val[N + 5], Son[2][N + 5], Level[N + 5], Size[N + 5], Num[N + 5];
inline void up(int u)
{
Size[u] = Size[Son[0][u]] + Size[Son[1][u]] + Num[u];
}
inline void Lturn(int &x)
{
int y = Son[1][x]; Son[1][x] = Son[0][y]; Son[0][y] = x;
up(x), up(y); x = y;
}
inline void Rturn(int &x)
{
int y = Son[0][x]; Son[0][x] = Son[1][y]; Son[1][y] = x;
up(x), up(y); x = y;
}
inline void Newnode(int &u, int v)
{
u = ++point;
Level[u] = rand(), Val[u] = v;
Num[u] = Size[u] = 1, Son[0][u] = Son[1][u] = 0;
}
void Insert(int &u, int t)
{
if (!u)
{
Newnode(u, t);
return;
}
Size[u]++;
if (t == Val[u]) Num[u]++;
else if (t < Val[u])
{
Insert(Son[0][u], t);
if (Level[Son[0][u]] < Level[u]) Rturn(u);
}
else
{
Insert(Son[1][u], t);
if (Level[Son[1][u]] < Level[u]) Lturn(u);
}
}
void Delete(int &u, int t)
{
if (!u) return;
if (Val[u] == t)
{
if (Num[u] > 1)
{
Size[u]--, Num[u]--;
return;
}
if (Son[0][u] * Son[1][u] == 0) u = Son[0][u] + Son[1][u];
else if (Level[Son[0][u]] < Level[Son[1][u]]) Rturn(u), Delete(u, t);
else Lturn(u), Delete(u, t);
}
else if (t < Val[u]) Size[u]--, Delete(Son[0][u], t);
else Size[u]--, Delete(Son[1][u], t);
}
int Find_num(int u, int t)
{
if (!u) return 0;
if (t <= Size[Son[0][u]]) return Find_num(Son[0][u], t);
else if (t <= Size[Son[0][u]] + Num[u]) return u;
else return Find_num(Son[1][u], t - Size[Son[0][u]] - Num[u]);
}
}Treap;
int sqz()
{
int n = read(), m = read();
rep(i, 1, n) T[i] = read();
rep(i, 1, m) Q[i].left = read(), Q[i].right = read(), Q[i].ask = read(), Q[i].id = i;
sort(Q + 1, Q + m + 1, cmp);
rep(i, Q[1].left, Q[1].right) Treap.Insert(root, T[i]);
ans[Q[1].id] = Treap.Val[Treap.Find_num(root, Q[1].ask)];
rep(i, 2, m)
{
if (Q[i].left <= Q[i - 1].right)
{
rep(j, Q[i - 1].left, Q[i].left - 1) Treap.Delete(root, T[j]);
rep(j, Q[i - 1].right + 1, Q[i].right) Treap.Insert(root, T[j]);
}
else
{
rep(j, Q[i - 1].left, Q[i - 1].right) Treap.Delete(root, T[j]);
rep(j, Q[i].left, Q[i].right) Treap.Insert(root, T[j]);
}
ans[Q[i].id] = Treap.Val[Treap.Find_num(root, Q[i].ask)];
}
rep(i, 1, m) printf("%d\n", ans[i]);
return 0;
}
