poj 2104 K-th Number(主席树)

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本文介绍如何使用主席树解决区间第K大数的问题，包括输入输出样例及核心代码实现，通过更新节点和查询区间信息来快速定位答案。

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 62606		Accepted: 22014
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 10 ⁹ 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

题意：

给了n个数,m次操作，每次操作询问x到y区间第k大的数(从小到大排序)。

思路：

第一次敲主席树，不太熟练。参考的B站里的这个代码。

https://www.bilibili.com/video/av4619406/?from=search&seid=17486694591577279544#page=2

讲一下自己对主席树的理解，主席树就是保留n个线段树的版本，每次更新的时候，就是把需要更新的节点就新建节点，而不需要改变的节点就保留不变。而且主席树也是前缀和的线段树，用tr[y]-tr[x-1]就可以得到x到y这个区间的信息。非常的方便。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn=1e5+5;
int cnt,root[maxn],n,m,a[maxn];
struct node
{
    int l,r,sum;
}tr[maxn*40];
vector<int>v;
int getid(int x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
void update(int l,int r,int &x,int y,int pos)
{
    tr[++cnt]=tr[y];
    tr[cnt].sum++;
    x=cnt;
    if(l==r) return;
    int mid=(l+r)/2;
    if(pos<=mid) update(l,mid,tr[x].l,tr[y].l,pos);
    else update(mid+1,r,tr[x].r,tr[y].r,pos);
}
int query(int l,int r,int x,int y,int k)
{
    if(l==r) return l;
    int sum=tr[tr[y].l].sum-tr[tr[x].l].sum;
    int mid=(l+r)/2;
    if(sum>=k) return query(l,mid,tr[x].l,tr[y].l,k);
    else return query(mid+1,r,tr[x].r,tr[y].r,k-sum);
}
int main()
{
    int x,y,k;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        v.push_back(a[i]);
    }
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
    for(int i=1;i<=n;i++)
    {
        update(1,n,root[i],root[i-1],getid(a[i]));
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&k);
        printf("%d\n",v[query(1,n,root[x-1],root[y],k)-1]);
    }
    return 0;
}