POJ 2104 K-th Number 区间第K大，可持久化线段树

Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1…n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: “What would be the k-th number in a[i…j] segment, if this segment was sorted?”
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2…5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input
The first line of the input file contains n — the size of the array, and m — the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 109 by their absolute values — the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output
For each question output the answer to it — the k-th number in sorted a[i…j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

题意就是求l,r区间的第k大是什么，学习了可持久化激动来yy此题，可持久化去年就开始看了，但是到今天也忘得很多了，最近重新YY了一下可持久化，感觉算是理解得比较清楚了，再次写这个题目。主席树的学习可以看这篇博客，写得很不错。
我也是照着模拟弄清楚了

#include <stdio.h>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 1e5+6;
int n, m, cnt, root[maxn], a[maxn], x, y, k;
struct node{int l, r, sum; } T[maxn*40];
vector <int> v;
int getid(int x){return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;}
void update(int l, int r, int &x, int y, int pos){
    T[++cnt] = T[y], T[cnt].sum++, x = cnt;
    if(l == r) return;
    int mid = (l + r)/2;
    if(pos <= mid) update(l, mid, T[x].l, T[y].l, pos);
    else update(mid+1, r, T[x].r, T[y].r, pos);
}
int query(int l, int r, int x, int y, int k){
    if(l == r) return l;
    int mid = (l + r)/2;
    int sum = T[T[y].l].sum - T[T[x].l].sum;
    if(sum >= k) return query(l, mid, T[x].l, T[y].l, k);
    else return query(mid+1, r, T[x].r, T[y].r, k - sum);
}
int main(){
    scanf("%d%d", &n,&m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), v.push_back(a[i]);
    sort(v.begin(), v.end()), v.erase(unique(v.begin(), v.end()), v.end());
    for(int i = 1; i <= n; i++) update(1, n, root[i], root[i-1], getid(a[i]));
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d", &x, &y, &k);
        printf("%d\n", v[query(1, n, root[x-1], root[y], k) - 1]);
    }
    return 0;
}