树套树

树状数组套平衡树

可以解决区间的可减问题， 比如区间有多少个大于等于/小于等于 k 的数。

对于 [这道题]， 其实也可以用树状数组套平衡树， 不过要外层二分多 log， 平衡树比较难多树二分。

具体地， 对于修改， 拆成插入删除；对于排名直接做；对于前驱后继 kth， 都可以用外层套二分做。

整体的复杂度， 空间是 O(q log n)， 时间是 O(q log w log² n) 的。

具体说说前驱后继二分的细节， 对于一个数 x 的前驱， 是 [0,x-1] 中最大的和不同的 x “大于等于其的有多少” 的数， 后继也差不多， 比较糊地说一下就是 x 往后找后第一个和 x 的 “小于等于其的有多少” 的数值不同的数。

#include<bits/stdc++.h>
#define INF 1e9
using namespace std;

const int N = 5e4 + 3, SZ = (5e4+23) * 2 * 16 + 2333;

int cnt;
#define newnode(s, v, a, b) (&(*pool[cnt++] = node(s, v, a, b)))
#define merge(a, b) newnode(a->siz+b->siz, b->val, a, b)
#define upd(me) if(me->ls->siz) me->siz=me->ls->siz+me->rs->siz, me->val=me->rs->val
struct node{
	int siz, val;
	node *ls, *rs;
	node(int s, int v, node *a, node *b) : siz(s), val(v), ls(a), rs(b) {
	}
	node() {
	}
} *root[N], *pool[SZ], t[SZ], *emp;

inline void maintain(node *me) {
	if(me->ls->siz > me->rs->siz * 4)
		me->rs = merge(me->ls->rs, me->rs), pool[--cnt] = me->ls, me->ls = me->ls->ls;
	if(me->rs->siz > me->ls->siz * 4)
		me->ls = merge(me->ls, me->rs->ls), pool[--cnt] = me->rs, me->rs = me->rs->rs;
}

void ins(int x, node *me) {
	if(me->siz == 1)
		me->ls = newnode(1, min(x, me->val), emp, emp),
		me->rs = newnode(1, max(x, me->val), emp, emp);
	else
		ins(x, x>me->ls->val ? me->rs : me->ls), maintain(me);
	upd(me);
}

void del(int x, node *me) {
	if(me->ls->siz == 1 && me->ls->val == x)
		pool[--cnt] = me->ls, pool[--cnt] = me->rs, *me = *me->rs;
	else if(me->rs->siz == 1 && me->rs->val == x)
		pool[--cnt] = me->ls, pool[--cnt] = me->rs, *me = *me->ls;
	else
		del(x, x>me->ls->val ? me->rs : me->ls), maintain(me);
	upd(me);
}

int low(int x, node *me) {
	if(me->siz == 1) return (me->val <= x);
	else return x>=me->ls->val ? me->ls->siz + low(x, me->rs) : low(x, me->ls);
} 

int upp(int x, node *me) {
	return me->siz - low(x-1, me);
}

int n, m, a[N];

void ins(int x, int v) {
	for(; x<=n; x += (x&(-x))) ins(v, root[x]);
}
void del(int x, int v) {
	for(; x<=n; x += (x&(-x))) del(v, root[x]);
}
int low(int x, int v) {
	int res = 0;
	for(; x; x-=(x&(-x))) res += low(v, root[x]);
	return res;
}
int low(int l, int r, int v) {
	return low(r, v) - low(l-1, v);
}
int upp(int x, int v) {
	int res = 0;
	for(; x; x-=(x&(-x))) res += upp(v, root[x]);
	return res;
}
int upp(int l, int r, int v) {
	return upp(r, v) - upp(l-1, v);
}

int main() {
	emp = new node(0, 0, NULL, NULL);
	for(int i=0; i<SZ; ++i) pool[i] = &t[i];
	scanf("%d%d", &n, &m);
	for(int i=1; i<=n; ++i) root[i] = newnode(1, INF, emp, emp);
	for(int i=1; i<=n; ++i) scanf("%d", &a[i]), ins(i, a[i]);
	while(m--)
	{
		int opt;
		scanf("%d", &opt);
		if(opt == 1)
		{
			int l, r, k;
			scanf("%d%d%d", &l, &r, &k);
			cout << low(l, r, k-1) + 1 << '\n';
		}
		if(opt == 2)
		{
			int l, r, k;
			scanf("%d%d%d", &l, &r, &k);
			int L=0, R=100000000;
			while(L != R)
			{
				int mid = (L+R+1) >> 1;
				if(low(l, r, mid-1) + 1 <= k) L = mid;
				else R = mid-1;
			}
			cout << L << '\n';
		}
		if(opt == 3)
		{
			int pos, k;
			scanf("%d%d", &pos, &k);
			del(pos, a[pos]);
			ins(pos, (a[pos] = k));
		}
		if(opt == 4)
		{
			int l, r, k;
			scanf("%d%d%d", &l, &r, &k);
			if(k == 0) {
				puts("-2147483647");
				continue;
			}
			int L=-1, R=k-1, tmp = upp(l, r, k);
			while(L!=R) {
				int mid = (L+R+1) >> 1;
				if(upp(l, r, mid) != tmp) L = mid;
				else R = mid-1;
			}
			
			if(L == -1) puts("-2147483647");
			else cout << L << '\n';
		}
		if(opt == 5)
		{
			int l, r, k;
			scanf("%d%d%d", &l, &r, &k);
			if(k == 100000000) {
				puts("2147483647");
				continue;
			}
			int L=k+1, R=100000000+1, tmp = low(l, r, k);
			while(L!=R) {
				int mid = (L+R) >> 1;
				if(low(l, r, mid) != tmp) R = mid;
				else L = mid+1;
			}
			
			if(L == 100000001) puts("2147483647");
			else cout << L << '\n';
		}
	}
	return 0;
}