HDU 2665 Kth number——————区间第k大，主席树

Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21499 Accepted Submission(s): 6560

Problem Description
Give you a sequence and ask you the $k_{th}$ 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<cstdio>
#include<vector>
#include<iostream>
#include<algorithm>

using namespace std;

const int MAXN = 1e5+7;
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[y] 是上一版本的， T[x] 是这一版本的
        T[++cnt] = T[y]; T[cnt].sum++; x = cnt;
        if(l == r) return ;
        int mid = (l+r)>>1;
        if(mid >= pos)  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)>>1;
        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() {
        int t;scanf("%d",&t);
        while(t--) {
                v.clear();
                cnt = 0;
                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;
}