hdoj 2665 Kth number主席树裸

Posted 2020-08-13

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9417    Accepted Submission(s): 2938

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

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 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 using namespace std;
 5 
 6 const int M = 1000008;
 7 
 8 struct Node{
 9     int r, l, sum;
10 }T[M*20];
11 struct A{
12     int idx, x;
13     bool operator <(const A & o)const{
14         return x < o.x;
15     }
16 }a[M];
17 int T_cnt;
18 void inser(int &num, int &x, int L, int R){
19     T[T_cnt++] = T[x]; x = T_cnt-1;
20     ++T[x].sum;
21     if(L == R) return;
22     int mid = (L+R)>>1;
23     if(num <= mid) inser(num, T[x].l, L, mid);
24     else inser(num, T[x].r, mid+1, R);
25 }
26 int query(int i, int j, int k, int L, int R){
27     if(L == R) return L;
28     int t = T[T[j].l].sum - T[T[i].l].sum;
29     int mid = (R+L)>>1;
30     if(k <= t) return query(T[i].l, T[j].l, k, L, mid);
31     else return query(T[i].r, T[j].r, k-t, mid+1, R);
32 }
33 int ranks[M], root[M];
34 int n, m, t;
35 int main(){
36     T[0].r = T[0].r = T[0].sum = 0;
37     root[0] = 0;
38     scanf("%d", &t);
39     while(t--){
40         scanf("%d%d", &n, &m);
41         for(int i = 1; i <= n; i++) {
42             scanf("%d", &a[i].x);
43             a[i].idx = i;
44         }
45         sort(a+1, a+n+1);
46         for(int i = 1; i <= n; i++){
47             ranks[a[i].idx] = i;
48         }
49         T_cnt = 1;
50         for(int i = 1; i <= n; i++){
51             root[i] = root[i-1];
52             inser(ranks[i], root[i], 1, n);
53         }
54         while(m--){
55             int i, j, k;
56             scanf("%d%d%d", &i, &j, &k);
57             printf("%d\n", a[query(root[i-1], root[j], k, 1, n)].x);
58         }
59     }
60 }