poj2104(划分树)

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 32718		Accepted: 10250
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

       本题首先给定一序列，要求给定查询区间内序列中的第k小值。

       关于区间查询问题，很容易想到线段树。但是本题是要求给定区间的第k小值，线段树并不能很好地解决，线段树适合处理RMQ问题。划分树算法思想参考http://blog.csdn.net/xj2419174554/article/details/10614203

贴个代码，是参考大牛博客http://www.cnblogs.com/kuangbin/category/404860.html的模板

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int MAXN=100000+100;
const int MAXPOW=20;//MAXPOW依据MAXN定大小
int tree[MAXPOW][MAXN];//tree[dep][i]表示第dep层第i个位置的值
int sorted[MAXN];//已经排序的数
int toleft[MAXPOW][MAXN];//toleft[dep][i]表示第dep层从1到i进入左边元素的个数

//构建深度为dep、区间为[l,r]的划分树
//时间复杂度为O(N*log(N))
void build(int l,int r,int dep)
{
	int i;
    if(l==r)return;
    int mid=(l+r)>>1;
    int same=mid-l+1;//表示等于中间元素而且被进入左边的元素个数
    for(i=l;i<=r;i++)
      if(tree[dep][i]<sorted[mid])
         same--;
    int lpos=l;
    int rpos=mid+1;
    for(i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])//比中间的元素小，进入左边
             tree[dep+1][lpos++]=tree[dep][i];
        else if(tree[dep][i]==sorted[mid]&&same>0)
        {
            tree[dep+1][lpos++]=tree[dep][i];
            same--;
        }
        else  //比中间的元素大，进入右边
            tree[dep+1][rpos++]=tree[dep][i];
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的元素个数
    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}


//查询小区间[l,r]内的第k大的元素，[L,R]是覆盖小区间的大区间
//时间复杂度为O(log(N))
int query(int L,int R,int l,int r,int dep,int k)
{
    if(l==r)return tree[dep][l];
    int mid=(L+R)>>1;
    int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的元素个数
    if(cnt>=k)//进入左子树查询
    {
        //修改小区间的l=L+要查询的区间前进入左边的元素个数
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        //r=newl+查询区间会被放在左边的元素个数
        int newr=newl+cnt-1;
        return query(L,mid,newl,newr,dep+1,k);
    }
    else//进入右子树查询
    {
		//修改小区间的r=r+要查询的区间后进入左边的元素个数
         int newr=r+toleft[dep][R]-toleft[dep][r];
		 //l=r-要查询的区间进入右边的元素个数
         int newl=newr-(r-l-cnt);
         return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}


int main()
{
    int n,m,i;
    int l,r,k;
    while(~scanf("%d%d",&n,&m))
    {
        memset(tree,0,sizeof(tree));//这个必须
        for(i=1;i<=n;i++)//从1开始
        {
            scanf("%d",&tree[0][i]);
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
        while(m--)
        {
            scanf("%d%d%d",&l,&r,&k);
            printf("%d\n",query(1,n,l,r,0,k));
        }
    }
    return 0;
}