HDU-2665-Kth number(划分树)_abc 295 e - kth number-优快云博客

本文链接：https://blog.youkuaiyun.com/faithdmc/article/details/37582867

本文介绍了一种解决区间K大数查询问题的有效算法。通过使用划分树的方法，该算法能够在给定序列中快速找到指定区间内的第K大的数值。文章详细解释了算法的实现过程，并提供了完整的C++代码示例。

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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

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思路：划分树板子题。

#include <cstdio>
#include <algorithm>
using namespace std;

int n,num[100005],sorted[100005],node[20][100005],sum[20][100005];

void build(int c,int s,int e)//第c层，s到e
{
    int mid,lm,lp,rp,i;

    mid=(s+e)>>1;
    lm=0;
    lp=s;//左子树的头指针
    rp=mid+1;//右子树的头指针

    for(i=s;i<=mid;i++) if(sorted[i]==sorted[mid]) lm++;//求出s到e有多少个数等于sorted[mid]

    for(i=s;i<=e;i++)
    {
        if(i==s) sum[c][i]=0;//计算出s到i之间有多少个被分到了左子树
        else sum[c][i]=sum[c][i-1];

        if(node[c][i]==sorted[mid])
        {
            if(lm)
            {
                lm--;
                sum[c][i]++;
                node[c+1][lp++]=node[c][i];
            }
            else node[c+1][rp++]=node[c][i];
        }
        else if(node[c][i]<sorted[mid])
        {
            sum[c][i]++;
            node[c+1][lp++]=node[c][i];
        }
        else node[c+1][rp++]=node[c][i];
    }

    if(s!=e)
    {
        build(c+1,s,mid);
        build(c+1,mid+1,e);
    }
}

int query(int c,int s,int e,int l,int r,int k)
{
    if(s==e) return node[c][s];
    else
    {
        int ls,rs,mid;

        mid=(s+e)>>1;

        if(s==l)//特判
        {
            ls=0;
            rs=sum[c][r];
        }
        else
        {
            ls=sum[c][l-1];
            rs=sum[c][r]-ls;
        }

        if(rs>=k) return query(c+1,s,mid,s+ls,s+ls+rs-1,k);//要查询的数在左子树
        else return query(c+1,mid+1,e,mid-s+1+l-ls,mid-s+1+r-ls-rs,k-rs);//要查询的数在右子树
    }
}

int main()
{
    int T,m,i,a,b,k;

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d%d",&n,&m);

        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);

            node[0][i]=sorted[i]=num[i];
        }

        sort(sorted+1,sorted+n+1);

        build(0,1,n);

        while(m--)
        {
            scanf("%d%d%d",&a,&b,&k);

            printf("%d\n",query(0,1,n,a,b,k));
        }
    }
}