POJ-2104 K-th Number (主席树 不带修改区间第k大)

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 61874		Accepted: 21744
Case Time Limit: 2000MS

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.

#include <stdio.h>
#include <algorithm> 
using namespace std;    
const int maxn = 100010;  
const int INF = 1 << 30;
int a[maxn], b[maxn], rt[maxn], tot;
struct tree{
	int sum, lson, rson;
}c[maxn << 5];
int newNode(int sum, int lson, int rson){
	int id = ++tot;
	c[id].sum = sum;
	c[id].lson = lson;
	c[id].rson = rson;
	return id;
}
void insert(int& rt, int pre, int pos, int l, int r){
	rt = newNode(c[pre].sum + 1, c[pre].lson, c[pre].rson);
	if(l == r) return;
	int mid = l + r >> 1;
	if(pos <= mid) insert(c[rt].lson, c[pre].lson, pos, l, mid);
	else insert(c[rt].rson, c[pre].rson, pos, mid + 1, r);
}
int query(int Lrt, int Rrt, int l, int r, int k){
	if(l == r) return l;
	int mid = l + r >> 1;
	int sum = c[c[Rrt].lson].sum - c[c[Lrt].lson].sum;
	if(k <= sum) return query(c[Lrt].lson, c[Rrt].lson, l, mid, k);
	else return query(c[Lrt].rson, c[Rrt].rson, mid + 1, r, k - sum);
}
int main(){
	int n, m, l, r, k, pos, num;
	while(scanf("%d %d", &n, &m) != EOF){
		tot = rt[0] = 0;
		for(int i = 1; i <= n; ++i){
			scanf("%d", &a[i]);
			b[i] = a[i];
		}
		sort(b + 1, b + 1 + n);
		num = unique(b + 1, b + 1 + n) - b - 1;
		for(int i = 1; i <= n; ++i){
			pos = lower_bound(b + 1, b + num + 1, a[i]) - b;
			insert(rt[i], rt[i - 1], pos, 1, num);
		}
		while(m--){
			scanf("%d %d %d", &l, &r, &k);
			pos = query(rt[l - 1], rt[r], 1, num, k);
			printf("%d\n", b[pos]);
		}
	}
}    
  
/*
题意：
1e5个数，5e4次询问，每次询问区间第k大。

思路：
将1e5个数离散化后映射到值域线段树上，这样线段树可以类似于平衡树，对于单次询问
我们可以通过左右儿子的sz大小来找第k大的标号。对于多次询问我们需要用到主席树，记录
历史的某个时刻线段树上的sz。
我们按照数字出现的顺序建n棵线段树，对于没有修改到的结点，我们直接连到上一棵的结点，
对于修改到的，我们开辟新结点。这样对于区间[L,R]，我们只需要看第L-1棵和第R棵线段树上
的sz来求第k大的标号即可。第L-1棵记录的是L之前出现的数字，减掉后就是在[L,R]之间出现的。
*/