Kth number（划分树）

末蓝、 2022-08-01 14:57 122阅读 0赞

# Kth number #

**Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 6217    Accepted Submission(s): 1986**

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

Input

The first line is the number of the test cases.    
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.    
The second line contains n integers, describe the sequence.    
Each of following m lines contains three integers s, t, k.    
\[s, t\] indicates the interval and k indicates the kth big number in interval \[s, t\]

Output

For each test case, output m lines. Each line contains the kth big number.

Sample Input

1 
    10 1 
    1 4 2 3 5 6 7 8 9 0 
    1 3 2

Sample Output

Source

[HDU男生专场公开赛——赶在女生之前先过节（From WHU）][HDU_From WHU]

**划分树裸题，我也没看懂，嘿嘿！**

#include <iostream>
    #include <string.h>
    #include <stdio.h>
    #include <algorithm>
    #define maxn 100010
    #define mid ((l+r)>>1)
    using namespace std;
    int t[20][maxn],sum[20][maxn];
    int a[maxn],as[maxn];
    //以下为查找区间第k小划分树
    void build(int p,int l,int r)
    {
        int lm=0,i,ls=l,rs=mid+1;//lm表示应被放入左子树且与中位数相等的数有多少个，ls为左子树的起始位置，rs为右子树的起始位置
        for(i=mid;i>=l;i--) //求lm
        {
            if(as[i]==as[mid])
            lm++;
            else
            break;
        }
        for(i=l;i<=r;i++)
        {
            if(i==l)//这里要特殊讨论
            sum[p][i]=0;
            else
            sum[p][i]=sum[p][i-1];
            if(t[p][i]==as[mid])//若与中位数相等则判断是否应该被放入左子树
            {
                if(lm)
                {
                    lm--;
                    sum[p][i]++;
                    t[p+1][ls++]=t[p][i];
                }
                else
                t[p+1][rs++]=t[p][i];
            }
            else if(t[p][i]<as[mid])//查找区间第K大即为>
            {
                sum[p][i]++;
                t[p+1][ls++]=t[p][i];
            }
            else
            t[p+1][rs++]=t[p][i];
        }
        if(l==r)
        return;
        build(p+1,l,mid);
        build(p+1,mid+1,r);
    }
    int query(int p,int l,int r,int ql,int qr,int k)
    {
        int s,ss;//s表示l到ql-1的区间内放入左子树的个数，ss表示区间[ql,qr]被放入左子树的个数
        if(l==r)//找到所求的数
        return t[p][l];
        if(ql==l)
        s=0,ss=sum[p][qr];
        else
        s=sum[p][ql-1],ss=sum[p][qr]-s;
        if(k<=ss)//要找的数在左子树中
        return query(p+1,l,mid,l+s,l+sum[p][qr]-1,k);
        else//要找的数在右子树中
        return query(p+1,mid+1,r,mid+1-l+ql-s,mid+1-l+qr-sum[p][qr],k-ss);
    }
    int main()
    {
        int t1,n,m,i,a,b,c;
        cin>>t1;
        while(t1--)
        {
            cin>>n>>m;
            for(i=1;i<=n;i++)
            {
                cin>>as[i];
                t[0][i]=as[i];
            }
            sort(as+1,as+n+1);
            build(0,1,n);
            while(m--)
            {
                cin>>a>>b>>c;
                cout<<query(0,1,n,a,b,c)<<endl;
            }
        }
    }

[HDU_From WHU]: http://acm.hdu.edu.cn/search.php?field=problem&key=HDU%C4%D0%C9%FA%D7%A8%B3%A1%B9%AB%BF%AA%C8%FC%A1%AA%A1%AA%B8%CF%D4%DA%C5%AE%C9%FA%D6%AE%C7%B0%CF%C8%B9%FD%BD%DA%A3%A8From+WHU%A3%A9&source=1&searchmode=source