【模板】可持久化线段树 2（主席树）

题目链接https://www.luogu.com.cn/problem/P3834

代码

#include<iostream>
#include<string>
#include<map>
#include<set>
//#include<unordered_map>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<stack>
#include<cmath>
#include<fstream>
#define X first
#define Y second
#define best 131 
#define INF 0x3f3f3f3f
#define P pair<int,int>
#define ls p<<1
#define rs p<<1|1
//#define int long long
//#define int long long
//const ll lnf=0x3f3f3f3f3f3f3f3f;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps=1e-5;
const double pai=acos(-1.0);
const int N=2e5+10;
const int maxn=1e6+10;
const int mod=1000000007;
//const int d[4][2]={0,1,1,0,-1,0,0,-1};
//int day[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};
int n,q,m,cnt=0;
int a[N],b[N],root[N];
int sum[N<<5],L[N<<5],R[N<<5];
inline void build(int p,int l, int r)
{
        p=++cnt,sum[p]=0;
	if(l==r) return;
	int mid=(l+r)>>1;
	build(L[p],l,mid); 
	build(R[p],mid+1,r);
}
inline int update(int p,int l,int r,int pos)
{
    int rt=++cnt;
    L[rt]=L[p],R[rt]=R[p];sum[rt]=sum[p]+1;
    if(l<r) 
    {
		int mid=(l+r)>>1;
        if(pos<=mid) L[rt]=update(L[p],l,mid,pos);
        else R[rt]=update(R[p],mid+1,r,pos);
    }
    return rt;
}
inline int query(int u, int v, int l, int r, int k)
{
    if(l>=r) return l;
    int mid=(l+r)>>1,x=sum[L[v]]-sum[L[u]];
    if(x>=k) return query(L[u],L[v],l,mid,k);
    else return query(R[u],R[v],mid+1,r,k-x);
}
int main()
{
    scanf("%d%d",&n,&q);
    for (int i=1;i<=n;i++)   
    {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    sort(b+1,b+1+n);
    int len=unique(b+1,b+1+n)-b-1;
    build(root[0],1,len);
    for(int i=1;i<=n;i++)  
    {
        int pos=lower_bound(b+1,b+1+len,a[i])-b;
        root[i]=update(root[i-1],1,len,pos);
    }
    while(q--)
    {
        int l,r,k;
        scanf("%d%d%d",&l,&r,&k);
        int pos=query(root[l-1],root[r],1,len,k);
        printf("%d\n",b[pos]);
    }
    return 0;
}