zoukankan      html  css  js  c++  java
  • POJ 2104 K-th Number 主席树

    K-th Number
     

    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.

     

    题意

      给你n个数,q次询问

      每次询问l,r之间的第k小的数是多少

    题解:

      不理解就去套板子啊

     

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int N = 1e5+10, M = 1e5, mod = 1e9, inf = 1e9+9;
    typedef long long ll;
    
    int n,q,a[N],pos[N],c[N];
    struct cooltree{
        struct node {int l,r,v;}tr[N*20];
        int sz,root[N];
        void init() {
            sz = 0, root[0] = 0;
        }
        void update(int &k,int l,int r,int x) {
            tr[++sz] = tr[k], k = sz;
            ++tr[k].v;
            if(l==r) return ;
            int mid = (l+r)>>1;
            if(x <= mid) update(tr[k].l,l,mid,x);
            else update(tr[k].r,mid+1,r,x);
        }
        int ask(int x,int y,int k) {
            int l = 1, r = n;
            x = root[x-1], y = root[y];
            while(l!=r) {
                int mid = (l+r)>>1, now = tr[tr[y].l].v - tr[tr[x].l].v;
                if(k <= now) x = tr[x].l,y = tr[y].l, r = mid;
                else x =tr[x].r , y = tr[y].r, l = mid + 1, k-=now;
            }
            return a[l];
        }
    }T;
    int main() {
        T.init();
        scanf("%d%d",&n,&q);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]), c[i] = a[i];
        sort(a+1,a+n+1);
        int cnt = unique(a+1,a+n+1) - a - 1;
        //cout<<cnt<<endl;
        for(int i=1;i<=n;i++) pos[i] = lower_bound(a+1,a+cnt+1,c[i]) - a;
        for(int i=1;i<=n;i++) T.update(T.root[i] = T.root[i-1],1,n,pos[i]);
        for(int i=1;i<=q;i++) {
                int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            printf("%d
    ",T.ask(x,y,z));
        }
        return 0;
    }
  • 相关阅读:
    深入Android 【一】 —— 序及开篇
    Android中ContentProvider和ContentResolver使用入门
    深入Android 【六】 —— 界面构造
    The service cannot be activated because it does not support ASP.NET compatibility. ASP.NET compatibility is enabled for this application. Turn off ASP.NET compatibility mode in the web.config or add the AspNetCompatibilityRequirements attribute to the ser
    Dynamic Business代码片段总结
    对文件的BuildAction以content,resource两种方式的读取
    paraview 3.12.0 windows下编译成功 小记
    百度网盘PanDownload使用Aria2满速下载
    netdata的安装与使用
    用PS给证件照排版教程
  • 原文地址:https://www.cnblogs.com/zxhl/p/5388532.html
Copyright © 2011-2022 走看看