zoukankan      html  css  js  c++  java
  • 主席树:POJ2104 K-th Number (主席树模板题)

    K-th Number
    Time Limit: 20000MS   Memory Limit: 65536K
    Total Submissions: 44952   Accepted: 14951
    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.
     
      这里要求序列的区间第K大,没有修改操作,于是作主席树模板打了~~~
      引用主席树介绍:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <algorithm>
     4 using namespace std;
     5 struct Node{
     6     int a,b,rs,ls,sum;
     7 }tr[2000010];
     8 int a[100010],b[100010];
     9 int rt[100010],pos,cnt;
    10 void Build(int &node,int a,int b)
    11 {
    12     node=++cnt;
    13     tr[node].a=a;
    14     tr[node].b=b;
    15     if(a==b)return;
    16     int mid=(a+b)>>1;
    17     Build(tr[node].ls,a,mid);
    18     Build(tr[node].rs,mid+1,b);
    19 }
    20 
    21 void Insert(int pre,int &node)
    22 {
    23     node=++cnt;
    24     tr[node].ls=tr[pre].ls;
    25     tr[node].rs=tr[pre].rs;
    26     tr[node].a=tr[pre].a;
    27     tr[node].b=tr[pre].b;
    28     tr[node].sum=tr[pre].sum+1;
    29     if(tr[node].a==tr[node].b)return;
    30     int mid=(tr[node].a+tr[node].b)>>1;
    31     if(mid>=pos)Insert(tr[pre].ls,tr[node].ls);
    32     else Insert(tr[pre].rs,tr[node].rs);
    33 }
    34 int Query(int pre,int node,int k)
    35 {
    36     if(tr[node].ls==tr[node].rs)return b[tr[node].a];
    37     int cmp=tr[tr[node].ls].sum-tr[tr[pre].ls].sum;
    38     if(cmp>=k)return Query(tr[pre].ls,tr[node].ls,k);
    39     else return Query(tr[pre].rs,tr[node].rs,k-cmp);
    40 }
    41 int main()
    42 {
    43     int n,q;
    44     scanf("%d%d",&n,&q);
    45     for(int i=1;i<=n;b[i]=a[i],i++)
    46         scanf("%d",&a[i]);
    47     sort(b+1,b+n+1);
    48     Build(rt[0],1,n);
    49     for(int i=1;i<=n;i++)
    50     {
    51         pos=lower_bound(b+1,b+n+1,a[i])-b;
    52         Insert(rt[i-1],rt[i]);
    53     }    
    54     int l,r,k;
    55     for(int i=1;i<=q;i++)
    56     {
    57         scanf("%d%d%d",&l,&r,&k);
    58         printf("%d
    ",Query(rt[l-1],rt[r],k));
    59     }
    60     return 0;
    61 }
    尽最大的努力,做最好的自己!
  • 相关阅读:
    Struts tags--Data tags
    Java NIO学习笔记七 Non-blocking Server
    Java NIO学习笔记六 SocketChannel 和 ServerSocketChannel
    Java NIO学习笔记五 FileChannel(文件通道)
    Java NIO学习笔记四 NIO选择器
    Java NIO学习笔记 三 散点/收集 和频道转换
    SpringMVC接收集合页面参数
    JAVA NIO学习笔记二 频道和缓冲区
    Java NIO学习笔记一 Java NIO概述
    通过举例了解java中的流
  • 原文地址:https://www.cnblogs.com/TenderRun/p/5202042.html
Copyright © 2011-2022 走看看