【POJ 2104】K-th Number

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 38927		Accepted: 12663
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 10 ⁹ 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

主席树求区间第k大，注释见代码。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <cstdlib>
#define maxn 100005
using namespace std;
int b[maxn],t[maxn],n,m,size,v[maxn],tot=0;
struct chairtree
{
	int l,r,size;
}a[maxn*30];
int Build(int l,int r)
{
	int root=++tot;
	a[root].size=0;
	if (l==r) return root;
	int m=(l+r)>>1;
	a[root].l=Build(l,m);
	a[root].r=Build(m+1,r);
	return root;
}
int Update(int root,int x)
{
	int now=++tot;
	int tmp=now;
	a[tot].size=a[root].size+1;
	int l=1,r=size;
	while (l<r)
	{
		int m=(l+r)>>1;
		if (x<=m)
		{
			a[now].l=++tot;
			a[now].r=a[root].r;   //充分利用原来相同的结点
			root=a[root].l;
			now=tot;
			r=m;
		}
		else
		{
			a[now].l=a[root].l;
			a[now].r=++tot;
			root=a[root].r;
			now=tot;
			l=m+1;
		}
		a[now].size=a[root].size+1;
	}
	return tmp;
}
int Ask(int lx,int rx,int k)
{
	int l=1,r=size;
	while (l<r)
	{
		int m=(l+r)>>1;
		if (a[a[rx].l].size-a[a[lx].l].size>=k)
		{
			r=m;
			lx=a[lx].l;
			rx=a[rx].l;
		}
		else
		{
			l=m+1;
			k-=a[a[rx].l].size-a[a[lx].l].size;
			lx=a[lx].r;
			rx=a[rx].r;
		}
	}
	return l;
}
void Hash1()
{
	sort(b+1,b+1+n);
	size=unique(b+1,b+1+n)-b-1;  //求不相同的数有几个
}
int Hash(int x)
{
	return lower_bound(b+1,b+1+size,x)-b;   //找到x在排好序的序列中出现的位置，即离散化后的值
}
int main()
{
    while (scanf("%d%d",&n,&m)==2)
	{
		for (int i=1;i<=n;i++)
			scanf("%d",&v[i]),b[i]=v[i];
		Hash1();
		t[0]=Build(1,size);
		for (int i=1;i<=n;i++)   //建n棵线段树
			t[i]=Update(t[i-1],Hash(v[i]));
		while (m--)
		{
			int l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			printf("%d\n",b[Ask(t[l-1],t[r],k)]);
		}
	}
	return 0;
}

注意：

1.unique函数中最后要-1，但还不知道原因。。以后再说。。

2.lower_bound中是+size而不是+all，记住。。