可持久化线段树【主席树】模板学习

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三道模板题（同求第k小）

//#include <bits/stdc++.h>
//#include <ext/pb_ds/priority_queue.hpp>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <algorithm>
//#pragma GCC optimize("O2")
using namespace std;
//using namespace __gnu_pbds;
#define Pair pair<LL,LL>
#define Combine Pair, greater<Pair>, pairing_heap_tag
#define LL long long
#define ll long long
#define ULL unsigned long long
#define ls rt<<1
#define rs rt<<1|1
#define one first
#define two second
#define MS 100009
#define INF 1e18
#define DBINF 1e100
#define Pi acos(-1.0)
#define eps 1e-9
#define mod 99999997

int n,m;
struct node{
	int count;
	int lls,rrs;
}p[MS<<5];
int rtpos[MS];
int a[MS],ta;
int b[MS];
int tot;

int build(int l,int r){
	int rt = ++tot;
	if(l == r) return rt;
	int m = l+r>>1;
	p[rt].lls = build(l,m);
	p[rt].rrs = build(m+1,r);
	return rt;
}

int update(int lart,int l,int r,int pos){
	int rt = ++tot;
	p[rt] = p[lart];
	p[rt].count++;
	if(l == r) return rt;
	int m = l+r>>1;
	if(m >= pos) p[rt].lls = update(p[lart].lls,l,m,pos);
	else p[rt].rrs = update(p[lart].rrs,m+1,r,pos);
	return rt;
}

int get_kth(int lrt,int rrt,int l,int r,int k){
	if(l == r) return l;
	int x = p[p[rrt].lls].count - p[p[lrt].lls].count;
	int m = l+r>>1;
	if(k <= x) return get_kth(p[lrt].lls,p[rrt].lls,l,m,k);
	else return get_kth(p[lrt].rrs,p[rrt].rrs,m+1,r,k-x);
}

void solve(){
	tot = 0;
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		b[i] = a[i];
	}
	sort(a+1,a+n+1);
	ta = 1;
	for(int i=2;i<=n;i++){
		if(a[i] != a[i-1]) a[++ta] = a[i];
	}
	rtpos[0] = build(1,ta);
	for(int i=1;i<=n;i++){
		int pos = lower_bound(a+1,a+ta+1,b[i]) - a;
		rtpos[i] = update(rtpos[i-1],1,ta,pos);
	}
	while(m--){
		int l,r,k;
		scanf("%d %d %d",&l,&r,&k);
		//k = r-l+1-k+1;
		int cc = get_kth(rtpos[l-1],rtpos[r],1,ta,k);
		printf("%d\n",a[cc]);
	}
}

int main() {
	//ios::sync_with_stdio(false);
	int ca;
	cin >> ca;
	while(ca--){
		solve();
	}

	return 0;

}