hdu 2665 Kth number(主席树模板)

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12299    Accepted Submission(s): 3730


Problem Description

Give you a sequence and ask you the kth big number of a inteval.

 

Input

The first line is the number of the test cases. 

For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere. 

The second line contains n integers, describe the sequence. 

Each of following m lines contains three integers s, t, k. 

[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

 

Output

For each test case, output m lines. Each line contains the kth big number.

 

Sample Input

1
10 1
1 4 2 3 5 6 7 8 9 0
1 3 2

 

Sample Output

2

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<map>
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5+100;
typedef long long LL;
int rt[N*20], ls[N*20], rs[N*20], sum[N*20];
int a
, b
, tot;
void build(int &o,int l,int r)
{
o= ++tot;
sum[o]=0;
if(l==r) return ;
int mid=(l+r)/2;
build(ls[o],l,mid);
build(rs[o],mid+1,r);
return ;
}
void update(int &o,int l,int r,int last,int p)
{
o= ++tot;
ls[o]=ls[last],rs[o]=rs[last];
sum[o]=sum[last]+1;
if(l==r) return ;
int mid=(l+r)/2;
if(p<=mid) update(ls[o],l,mid,ls[last],p);
else update(rs[o],mid+1,r,rs[last],p);
return ;
}
int query(int ss,int tt,int l,int r,int cnt)
{
if(l==r) return l;
int tmp=sum[ls[tt]]-sum[ls[ss]];
int mid=(l+r)/2;
if(tmp>=cnt) return query(ls[ss],ls[tt],l,mid,cnt);
else return query(rs[ss],rs[tt],mid+1,r,cnt-tmp);
}

int main()
{
int t, ncase=1;
scanf("%d", &t);
while(t--)
{
int n, q;
scanf("%d %d", &n, &q);
for(int i=1;i<=n;i++) scanf("%d", &a[i]), b[i]=a[i];
sort(b+1,b+n+1);
int k=unique(b+1,b+n+1)-(b+1);
tot=0;
build(rt[0],1,k);
for(int i=1;i<=n;i++)
{
int pos=lower_bound(b+1,b+k+1,a[i])-(b);
update(rt[i],1,k,rt[i-1],pos);
}
while(q--)
{
int l, r, x;
scanf("%d %d %d", &l, &r, &x);
printf("%d\n",b[query(rt[l-1],rt[r],1,k,x)]);
}
}
return 0;
}