POJ2104 K-th Number (子区间内第k大的数字)【划分树算法模板应用】

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 40920		Accepted: 13367
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.

算法及题目分析：给你一个具有n个数字的数组，进行m此询问。每次询问，都会给你一个子区间，问你这个区间内的第k大的数字是多少？例如：数组序列（下标从1开始）是：1 5 2 6 3 7 4，  询问：[2, 5]的第3大的数字。[2, 5]区间的数是：5 2 6 3，排序后：2 3 5 6，第3大的数应该是5. 但是当数据量比较庞大的时候，询问次数比较多的时候，这种常规算法思想实现起来之星效率比较低。于是一种树型结构算法应运而生：【划分树算法】（基于线段树思想）（划分树实现细节有待补充）。【可以参考：程序设计 解题策略 吴永辉 王建德 编著 P2】

代码：

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <ctype.h>

#include <math.h>

#include <string>

#include <iostream>

#include <algorithm>

#define N 100000+10



using namespace std;



//POJ 2104

int tree[50][N];//tree[p][i]表示第p层中第i的位置的值

int sorted[N];

int toleft[50][N];//toleft[p][i]表示第p层从1到i中有

//多少个数被划分入下一层的左子区间



//划分树建树的实现过程

void build(int ll ,int r, int dep)

{

    if(ll==r)

        return; //递归出口 若划分至叶子，则回溯

    int mid=(ll+r)>>1; //等于(l+r)/2 计算区间的中间指针

    int same=mid-ll+1; //计算[l,r]被分入下层左区间的个数

    for(int i=ll; i<=r; i++)

    {

        if(tree[dep][i] < sorted[mid])

            same--;

    }

    int llpos=ll;

    int rpos=mid+1;

    for(int i=ll; i<=r; i++)

    {

        if(tree[dep][i] < sorted[mid])

            tree[dep+1][llpos++]=tree[dep][i];

        else if(tree[dep][i]==sorted[mid] && same>0 )

        {

            tree[dep+1][llpos++]=tree[dep][i];

            same--;

        }

        else

            tree[dep+1][rpos++]=tree[dep][i];

        toleft[dep][i]=toleft[dep][ll-1] +llpos-ll;

    }

    build(ll, mid, dep+1);//递归计算下一层的左子区间

    build(mid+1, r, dep+1);//递归计算下一层的右子区间

}



int query(int L, int R, int ll, int r, int dep, int k)

//从划分数的dep层出发 自上而下的在大区间[L,R]里查询子区间

//[ll,r]中第K大的数

{

    if(ll==r)

        return tree[dep][ll];

    int mid=(L+R)>>1;

    int cnt=toleft[dep][r]-toleft[dep][ll-1];

    if(cnt >= k)

    {

        int newll=L+toleft[dep][ll-1]-toleft[dep][L-1];

        int newr = newll+cnt-1;

        return query(L, mid, newll, newr, dep+1, k);

    }

    else

    {

        int newr=r+toleft[dep][R]-toleft[dep][r];

        int newll=newr-(r-ll-cnt);

        return query(mid+1, R, newll, newr, dep+1, k-cnt);

    }

}



int main()

{

    int n, m;

    int i;



    while(~scanf("%d %d", &n, &m))

    {

        for(i=1; i<=n; i++)//n

        {

            scanf("%d", &tree[0][i]);//

            sorted[i]=tree[0][i];

        }

        sort(sorted+1,sorted+n+1);

        build(1, n, 0);//建立划分树

        while(m--)

        {

            int u, v, w;

            scanf("%d %d %d", &u, &v, &w);

            printf("%d\n", query(1, n, u, v, 0, w));

        }

    }

    return 0;

}