POJ 2104 K-th Number (Chair tree), pojk-th

POJ 2104 K-th Number (Chair tree), pojk-th
K-th Number

Time Limit:20000 MS		Memory Limit:65536 K
Total Submissions:57427		Accepted:19856
Case Time Limit:2000 MS

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 wocould 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 wocould 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 shocould 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 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

563

Hint

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

Source

Northeastern Europe 2004, template Question of the Northern Subregion chair tree.

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <cmath> 5 # include <queue> 6 # include <algorithm> 7 using namespace std; 8 const int MAXN = 10000001; 9 void read (int & n) 10 {11 char c = '+'; int x = 0; bool flag = 0; 12 while (c <'0' | c> '9') 13 {c = getchar (); if (c = '-') flag = 1 ;} 14 while (c >='0' & c <= '9') 15 {x = x * 10 + c-48, c = getchar ();} 16 flag = 1? N =-x: n = x; 17} 18 int n, m; 19 int a [MAXN]; 20 int hash [MAXN]; 21 int root [MAXN/2]; 22 int now, tot; // total number of occurrences of all nodes 23 int ls [MAXN], rs [MAXN], cnt [MAXN]; 24 void build (int & cur, int l, int r) 25 {26 cur = tot ++; // total number of nodes 27 cnt [cur] = 0; 28 if (l! = R) 29 {30 int mid = (l + r)/2; 31 build (ls [cur], l, mid); 32 build (rs [cur], mid + 1, r); 33} 34} 35 void update (int pre, int pos, int & cur, int l, int r) 36 {37 cur = tot ++; 38 cnt [cur] = cnt [pre] + 1; 39 ls [cur] = ls [pre]; rs [cur] = rs [pre]; 40 if (l = r) 41 return; 42 int mid = (l + r)/2; 43 if (pos <= mid) 44 update (ls [pre], pos, ls [cur], l, mid); 45 else 46 update (rs [pre], pos, rs [cur], mid + 1, r ); 47} 48 int query (int lt, int rt, int l, int r, int k) 49 {50 if (l = r) 51 return l; 52 int now = cnt [ls [rt]-cnt [ls [lt]; 53 int mid = (l + r)/2; 54 if (k <= now) 55 return query (ls [lt], ls [rt], l, mid, k); 56 else 57 return query (rs [lt], rs [rt], mid + 1, r, k-now); 58} 59 int main () 60 {61 while (scanf ("% d", & n, & m) = 2) 62 {63 for (int I = 1; I <= n; I ++) 64 {65 read (a [I]); 66 hash [I] = a [I]; 67} 68 sort (hash + 1, hash + n + 1); 69 int size = unique (hash + 1, hash + n + 1)-hash-1; 70 for (int I = 1; I <= n; I ++) 71 a [I] = lower_bound (hash + 1, hash + size + 1, a [I])-hash; // sort + discretization 72 73 tot = 0; 74 build (root [0], 1, size ); 75 76 for (int I = 1; I <= n; I ++) 77 update (root [I-1], a [I], root [I], 1, size); // create all trees 78 79 for (int I = 1; I <= m; I ++) 80 {81 int x, y, z; 82 read (x); read (y); read (z); 83 printf ("% d \ n", hash [query (root [x-1], root [y], 1, size, z)]); 84} // query 85} 86 87 return 0; 88}