poj2104 K-th Number（主席树静态区间第k大）

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.

Source

Northeastern Europe 2004, Northern Subregion

 

 

题目大意：给出长度为n的序列和m个形如(i,j,k)的询问，询问区间[i,j]中第k大的数。

 

本题是经典的主席树（可持久化线段树）模板题，时间复杂度O(mlog2n)，空间复杂度O(nlog2n)。

program rrr(input,output);
type
  treetype=record
     lc,rc,s:longint;
  end;
var
  a:array[0..100010*30]of treetype;
  rot,idx,b,c:array[0..100010]of longint;
  n,m,i,cnt,l,r,j,k,mid,x,y,t:longint;
procedure sort(q,h:longint);
var
  i,j,x,t:longint;
begin
   i:=q;j:=h;x:=b[(i+j)>>1];
   repeat
     while b[i]<x do inc(i);
     while x<b[j] do dec(j);
     if i<=j then
        begin
           t:=b[i];b[i]:=b[j];b[j]:=t;
           t:=idx[i];idx[i]:=idx[j];idx[j]:=t;
           inc(i);dec(j);
        end;
   until i>j;
   if j>q then sort(q,j);
   if i<h then sort(i,h);
end;
procedure build(k,l,r:longint);
var
  mid:longint;
begin
   a[k].s:=0;
   if l=r then exit;
   mid:=(l+r)>>1;
   inc(cnt);a[k].lc:=cnt;build(cnt,l,mid);
   inc(cnt);a[k].rc:=cnt;build(cnt,mid+1,r);
end;
procedure add(x:longint);
begin
   inc(cnt);rot[i]:=cnt;
   j:=cnt;k:=rot[i-1];a[j].s:=a[k].s+1;
   l:=1;r:=n;
   while l<r do
      begin
         mid:=(l+r)>>1;inc(cnt);
         if x>mid then
            begin
               l:=mid+1;
               a[j].lc:=a[k].lc;k:=a[k].rc;
               a[j].rc:=cnt;j:=cnt;a[j].s:=a[k].s+1;
            end
         else
            begin
               r:=mid;
               a[j].rc:=a[k].rc;k:=a[k].lc;
               a[j].lc:=cnt;j:=cnt;a[j].s:=a[k].s+1;
            end;
      end;
end;
function ask(x,y,t:longint):longint;
begin
   j:=rot[x-1];k:=rot[y];l:=1;r:=n;cnt:=0;
   while l<r do
      begin
         mid:=(l+r)>>1;
         if cnt+a[a[k].lc].s-a[a[j].lc].s>=t then begin r:=mid;j:=a[j].lc;k:=a[k].lc; end
         else begin cnt:=cnt+a[a[k].lc].s-a[a[j].lc].s;l:=mid+1;j:=a[j].rc;k:=a[k].rc; end;
      end;
   exit(b[l]);
end;
begin
   assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
   readln(n,m);
   for i:=1 to n do begin read(b[i]);idx[i]:=i; end;
   sort(1,n);
   for i:=1 to n do c[idx[i]]:=i;
   cnt:=1;rot[0]:=1;
   build(1,1,n);
   for i:=1 to n do add(c[i]);
   //for i:=1 to 40 do writeln(a[rot[i]].s);
   for i:=1 to m do begin readln(x,y,t);writeln(ask(x,y,t)); end;
   close(input);close(output);
end.