K-th Number
Time Limit: 20000MS   Memory Limit: 65536K
Total Submissions: 52348   Accepted: 17985
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.

Source

Northeastern Europe 2004, Northern Subregion
 
 
思路:
  可持久化线段树模板;
 
来,上代码:
  结构体数组版(ac):

#include <cstdio>
#include <algorithm> using namespace std; struct TreeNode {
int l,r,dis;
}; class PersistentLineSegmentTree {
private:
int n,m,if_z,tot,num[],num_[],size; char Cget; TreeNode root[],node[*]; inline void read_int(int &now_)
{
now_=,if_z=,Cget=getchar();
while(Cget>''||Cget<'')
{
if(Cget=='-') if_z=-;
Cget=getchar();
}
while(Cget>=''&&Cget<='')
{
now_=now_*+Cget-'';
Cget=getchar();
}
now_*=if_z;
} public:
PersistentLineSegmentTree()
{
read_int(n),read_int(m);
for(int i=;i<=n;i++)
{
read_int(num[i]);
num_[i]=num[i];
}
sort(num+,num+n+);
size=unique(num+,num+n+)-num-;
Build(,size,root[]);
for(int i=;i<=n;i++)
{
num_[i]=lower_bound(num+,num+size+,num_[i])-num;
Down(,size,root[i-],root[i],num_[i]);
}
int lit,rit,k;
for(int i=;i<=m;i++)
{
read_int(lit),read_int(rit),read_int(k);
printf("%d\n",num[Query(,size,root[lit-],root[rit],k)]);
}
} inline void Up(TreeNode &now)
{
now.dis=node[now.l].dis+node[now.r].dis;
} void Build(int l,int r,TreeNode &now)
{
if(l==r) return ;
int mid=(l+r)>>;
now.l=++tot;
Build(l,mid,node[now.l]);
now.r=++tot;
Build(mid+,r,node[now.r]);
} void Down(int l,int r,TreeNode &pre,TreeNode &now,int to)
{
if(l==r)
{
now.dis++;
return ;
}
int mid=(l+r)>>;
if(to>mid)
{
now.l=pre.l;
now.r=++tot;
Down(mid+,r,node[pre.r],node[now.r],to);
}
else
{
now.r=pre.r;
now.l=++tot;
Down(l,mid,node[pre.l],node[now.l],to);
}
Up(now);
} int Query(int l,int r,TreeNode &pre,TreeNode &suc,int K)
{
if(l==r)
{
return l;
}
int mid=(l+r)>>;
int dis=node[suc.l].dis-node[pre.l].dis;
if(K>dis) return Query(mid+,r,node[pre.r],node[suc.r],K-dis);
else return Query(l,mid,node[pre.l],node[suc.l],K);
}
}; class PersistentLineSegmentTree tree; int main()
{
return ;
}

指针版(正确但超时):

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; struct TreeNode {
int l,r,mid,dis; TreeNode *left,*right;
}; class PersistentLineSegmentTree {
private:
int n,m,if_z,num[],num_[],size; char Cget; TreeNode *root[],*null; inline void read_int(int &now_)
{
now_=,if_z=,Cget=getchar();
while(Cget>''||Cget<'')
{
if(Cget=='-') if_z=-;
Cget=getchar();
}
while(Cget>=''&&Cget<='')
{
now_=now_*+Cget-'';
Cget=getchar();
}
now_*=if_z;
} public:
PersistentLineSegmentTree()
{
null=new TreeNode;
null->left=null;
null->right=null;
root[]=null;
read_int(n),read_int(m);
for(int i=;i<=n;i++)
{
read_int(num[i]);
num_[i]=num[i];
root[i]=null;
}
sort(num+,num+n+);
size=unique(num+,num+n+)-num-;
Build(root[],,size);
for(int i=;i<=n;i++)
{
num_[i]=lower_bound(num+,num+size+,num_[i])-num;
Down(root[i-],root[i],num_[i]);
}
int lit,rit,k;
for(int i=;i<=m;i++)
{
read_int(lit),read_int(rit),read_int(k);
printf("%d\n",num[Query(root[lit-],root[rit],k)]);
}
} void Build(TreeNode *&now,int l,int r)
{
if(now==null)
{
now=new TreeNode;
now->l=l;
now->r=r;
now->dis=;
now->mid=(l+r)>>;
now->left=null;
now->right=null;
}
if(l==r) return ;
Build(now->left,l,now->mid);
Build(now->right,now->mid+,r);
} inline void Up(TreeNode *&now)
{
now->dis=now->left->dis+now->right->dis;
} void Down(TreeNode *&pre,TreeNode *&now,int to)
{
now=new TreeNode;
now->l=pre->l;
now->r=pre->r;
now->mid=pre->mid;
now->left=null;
now->right=null;
if(now->l==now->r)
{
now->dis=pre->dis+;
return ;
}
if(to>now->mid)
{
now->left=pre->left;
Down(pre->right,now->right,to);
}
else
{
now->right=pre->right;
Down(pre->left,now->left,to);
}
Up(now);
} int Query(TreeNode *&pre,TreeNode *&now,int K)
{
if(now->l==now->r) return now->l;
int dis=now->left->dis-pre->left->dis;
if(dis<K) return Query(pre->right,now->right,K-dis);
else return Query(pre->left,now->left,K);
}
};
class PersistentLineSegmentTree tree; int main()
{
return ;
}

AC日记——K-th Number poj 2104的更多相关文章

  1. K-th Number POJ - 2104

    K-th Number POJ - 2104 You are working for Macrohard company in data structures department. After fa ...

  2. K-th Number Poj - 2104 主席树

    K-th Number Poj - 2104 主席树 题意 给你n数字,然后有m次询问,询问一段区间内的第k小的数. 解题思路 这个题是限时训练做的题,我不会,看到这个题我开始是拒绝的,虽然题意清晰简 ...

  3. 主席树 【权值线段树】 && 例题K-th Number POJ - 2104

    一.主席树与权值线段树区别 主席树是由许多权值线段树构成,单独的权值线段树只能解决寻找整个区间第k大/小值问题(什么叫整个区间,比如你对区间[1,8]建立一颗对应权值线段树,那么你不能询问区间[2,5 ...

  4. K-th Number POJ - 2104 划分树

    K-th Number You are working for Macrohard company in data structures department. After failing your ...

  5. AC日记——Destroying The Graph poj 2125

    Destroying The Graph Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8356   Accepted: 2 ...

  6. HDU 2665.Kth number-可持久化线段树(无修改区间第K小)模板 (POJ 2104.K-th Number 、洛谷 P3834 【模板】可持久化线段树 1(主席树)只是输入格式不一样,其他几乎都一样的)

    Kth number Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  7. AC日记——青蛙的约会 poj 1061

    青蛙的约会 POJ - 1061   思路: 扩展欧几里得: 设青蛙们要跳k步,我们可以得出式子 m*k+a≡n*k+b(mod l) 式子变形得到 m*k+a-n*k-b=t*l (m-n)*k-t ...

  8. hdu 2665 Kth number (poj 2104 K-th Number) 划分树

    划分树的基本功能是,对一个给定的数组,求区间[l,r]内的第k大(小)数. 划分树的基本思想是分治,每次查询复杂度为O(log(n)),n是数组规模. 具体原理见http://baike.baidu. ...

  9. poj 2104 K-th Number 主席树+超级详细解释

    poj 2104 K-th Number 主席树+超级详细解释 传送门:K-th Number 题目大意:给出一段数列,让你求[L,R]区间内第几大的数字! 在这里先介绍一下主席树! 如果想了解什么是 ...

随机推荐

  1. keil swd设置下载stm32f103c8t6.

    1.debug选项,选择jlink,2.utilities选择jlink3.加载flash算法.4.选择swd模式,其他基本上默认,这样就可以下载了对rom和ram设置需要说明一下:1,IROM1,前 ...

  2. 大数运算:HDU-1042-N!(附N!位数的计算)

    解题心得: 这里使用了10000进制.很明显,因为是n!所以单个最大的数是10000*10000,使用万进制. 可以借鉴高精度的加法,单个乘了之后在进位. 很坑的一点,0!=1,数学不好WA了三次,尴 ...

  3. [BZOJ2331]地板(插头DP)

    Description lxhgww的小名叫"小L",这是因为他总是很喜欢L型的东西.小L家的客厅是一个的矩形,现在他想用L型的地板来铺满整个客厅,客厅里有些位置有柱子,不能铺地板 ...

  4. MAX(数论)

    Description 小C有n个区间,其中第i个区间为[li,ri],小C想从每个区间中各选出一个整数,使得所有选出的数and起来得到的结果最大,请你求出这个值. Input Format 第一行一 ...

  5. adb offline解决办法

    假如你连接手机之后,adb devices找不到设备,或者找到了设备,但是device ID后总是offline的状态,那估计就是驱动有问题. 强烈建议1.安装豌豆荚,它可以自己主动修复手机驱动,一般 ...

  6. python项目中输出指定颜色的日志

    起因 在开发项目过程中,为了方便调试代码,经常会向stdout中输出一些日志,默认的这些日志就直接显示在了终端中.而一般的应用服务器,第三方库,甚至服务器的一些通告也会在终端中显示,这样就搅乱了我们想 ...

  7. P3386 【模板】二分图匹配(匈牙利&最大流)

    P3386 [模板]二分图匹配 题目背景 二分图 题目描述 给定一个二分图,结点个数分别为n,m,边数为e,求二分图最大匹配数 输入输出格式 输入格式: 第一行,n,m,e 第二至e+1行,每行两个正 ...

  8. session为什么需要持久化

    为什么需要持久化: 客户端访问了某个能开启会话功能的资源, web服务器就会创建一个与该客户端对应的HttpSession对象,每个HttpSession对象都要站用一定的内存空间.如果在某一时间段内 ...

  9. day20 Django Models 操作,多表,多对多

    1 Django models 获取数据的三种方式: 实践: viwes def business(request): v1 = models.Business.objects.all() v2 = ...

  10. 【NOIP 2017 普及组】 跳房子

    裸的单调队列优化dp+二分 我居然还调了挺久 日常审题错误 #include <bits/stdc++.h> using namespace std; typedef long long ...