PKU2104K-th Number

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本文深入探讨了K-thNumber算法的原理及其在数组排序中的应用，通过实例解析了如何快速找到数组段中第k小的数。文章还提供了算法实现的详细代码，包括数据结构设计和输入输出优化策略。

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K-th Number
Time Limit: 20000MS Memory Limit: 65536K
Total Submissions: 44766 Accepted: 14869
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
主席树模板题。。
一般学习了一种新的数据结构或算法，不都有一道模板题来练手么。。

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,size,tot=0,root[100001],a[100001],ls[8000001],rs[8000001],s[8000001],num[100001],hash[100001];

int find(int x)
{
    int l=1,r=tot,mid;
    while (l<=r)
      {
        mid=(l+r)/2;
        if (hash[mid]<x)
          l=mid+1;
        else
          r=mid-1;
      }
    return l;
}

void insert(int l,int r,int x,int &y,int v)
{
    int mid;
    size++;
    y=size;
    s[y]=s[x]+1;
    if (l==r)
      return;
    ls[y]=ls[x];
    rs[y]=rs[x];
    mid=(l+r)/2;
    if (v<=mid)
      insert(l,mid,ls[x],ls[y],v);
    else
      insert(mid+1,r,rs[x],rs[y],v);
}

int query(int l,int r,int x,int y,int k)
{
    int mid;
    if (l==r)
      return l;
    mid=(l+r)/2;
    if (s[ls[y]]-s[ls[x]]>=k)
      return query(l,mid,ls[x],ls[y],k);
    else
      return query(mid+1,r,rs[x],rs[y],k-(s[ls[y]]-s[ls[x]]));
}

int main()
{
    int l,r,x,i;
    scanf("%d %d",&n,&m);
    for (i=1;i<=n;i++)
      {
        scanf("%d",&a[i]);
        num[i]=a[i];
      }
    sort(num+1,num+n+1);
    tot++;
    hash[tot]=num[1];
    for (i=2;i<=n;i++)
      if (num[i]!=num[i-1])
        {
            tot++;
            hash[tot]=num[i];
        }
    for (i=1;i<=n;i++)
      insert(1,tot,root[i-1],root[i],find(a[i]));
    for (i=1;i<=m;i++)
      {
        scanf("%d %d %d",&l,&r,&x);
        printf("%d\n",hash[query(1,tot,root[l-1],root[r],x)]);
      }
    return 0;
}