zoukankan      html  css  js  c++  java
  • 主席树K-th Number

    /*K-th Number
    Time Limit: 20000MS Memory Limit: 65536K
    Total Submissions: 44535 Accepted: 14779
    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*/
    #include<cstdio>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int ls[8000008],rs[8000008],sum[8000008],root[100008],num[100009],n,m,hash[100009];
    int tmp,size,a[100008];
    void jia(int l,int r,int x,int &y,int v)
    {
    y=++size;
    sum[y]=sum[x]+1;
    if(l==r)
    return;
    ls[y]=ls[x];
    rs[y]=rs[x];
    int mid=(l+r)>>1;
    if(v<=mid)
    jia(l,mid,ls[x],ls[y],v);
    else
    jia(mid+1,r,rs[x],rs[y],v);
    return;
    }
    int xun(int L,int R,int V)
    {
    int x=root[L-1],y=root[R],l=1,r=tmp,mid=(l+r)/2;
    for(;l!=r;)
    if(sum[ls[y]]-sum[ls[x]]>=V)
    {
    x=ls[x];
    y=ls[y];
    r=mid;
    mid=(l+r)>>1;
    }
    else
    {
    V-=sum[ls[y]]-sum[ls[x]];
    x=rs[x];
    y=rs[y];
    l=mid+1;
    mid=(l+r)>>1;
    }
    return l;
    }
    int main()
    {
    scanf("%d%d",&n,&m);
    for(int i=0;i<n; i++)
    {
    scanf("%d",&num[i]);
    a[i+1]=num[i];
    }
    sort(num,num+n);
    hash[++tmp]=num[0];
    for(int i=1;i<n;i++)
    if(hash[tmp]!=num[i])
    hash[++tmp]=num[i];
    for(int i=1;i<=n;i++)
    jia(1,tmp,root[i-1],root[i],lower_bound(hash+1,hash+n+1,a[i])-hash);
    for(int i=0;i<m;i++)
    {
    int l,r,v;
    scanf("%d%d%d",&l,&r,&v);
    printf("%d ",hash[xun(l,r,v)]);
    }
    return 0;
    }

  • 相关阅读:
    到底如何设置 Java 线程池的大小?
    面试一个 3 年 Java 程序员,一个问题都不会!
    Spring Boot 集成 Ehcache 缓存,三步搞定!
    牛逼哄哄的 "零拷贝" 是什么?
    一个 Java 字符串到底有多少个字符?
    不用找了,300 分钟帮你搞定 Spring Cloud!
    五分钟搞懂 Linux 重点知识,傻瓜都能学会!
    如何设计一个完美的权限管理模块?
    Redis基础都不会,好意思出去面试?
    .net c# MVC提交表单的4种方法
  • 原文地址:https://www.cnblogs.com/xydddd/p/5144142.html
Copyright © 2011-2022 走看看