HDU 2665 Kth number（划分树入门题，纯套模板）

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2026    Accepted Submission(s): 672

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

2

 

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

 

Recommend

zty

 

 

 

划分树。模板》。。

 

/*

HDU  2665 Kth number

划分树





*/





#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;



const int MAXN=100010;

int tree[30][MAXN];//表示每层每个位置的值

int sorted[MAXN];//已经排序的数

int toleft[30][MAXN];//toleft[p][i]表示第i层从1到i有多少个数分入左边



void build(int l,int r,int dep)

{

    if(l==r)return;

    int mid=(l+r)>>1;

    int same=mid-l+1;//表示等于中间值而且被分入左边的个数

    for(int i=l;i<=r;i++)

      if(tree[dep][i]<sorted[mid])

         same--;

    int lpos=l;

    int rpos=mid+1;

    for(int i=l;i<=r;i++)

    {

        if(tree[dep][i]<sorted[mid])//比中间的数小，分入左边

             tree[dep+1][lpos++]=tree[dep][i];

        else if(tree[dep][i]==sorted[mid]&&same>0)

        {

            tree[dep+1][lpos++]=tree[dep][i];

            same--;

        }

        else  //比中间值大分入右边

            tree[dep+1][rpos++]=tree[dep][i];

        toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数



    }

    build(l,mid,dep+1);

    build(mid+1,r,dep+1);



}





//查询区间第k大的数，[L,R]是大区间，[l,r]是要查询的小区间

int query(int L,int R,int l,int r,int dep,int k)

{

    if(l==r)return tree[dep][l];

    int mid=(L+R)>>1;

    int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数

    if(cnt>=k)

    {

        //L+要查询的区间前被放在左边的个数

        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];

        //左端点加上查询区间会被放在左边的个数

        int newr=newl+cnt-1;

        return query(L,mid,newl,newr,dep+1,k);

    }

    else

    {

         int newr=r+toleft[dep][R]-toleft[dep][r];

         int newl=newr-(r-l-cnt);

         return query(mid+1,R,newl,newr,dep+1,k-cnt);

    }

}





int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int T;

    int n,m;

    int s,t,k;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        memset(tree,0,sizeof(tree));//这个必须

        for(int i=1;i<=n;i++)//从1开始

        {

            scanf("%d",&tree[0][i]);

            sorted[i]=tree[0][i];

        }

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

        build(1,n,0);

        while(m--)

        {

            scanf("%d%d%d",&s,&t,&k);

            printf("%d\n",query(1,n,s,t,0,k));

        }

    }

    return 0;

}