hdu 2665 划分树

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7616    Accepted Submission(s): 2402

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
   
   
   
   

题意：

给你一串数字，然后是m次查询，每次查找[s,t]区间的第k大值      /*划分树学习

/*
模板题，hdu2665
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 100010;

int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];


void build(int l,int r,int dep)  //模拟快排 并记录左树中比i小的个数
{
    if(l == r)
        return;
    int mid = (l+r)>>1;
    int same = mid-l+1;//可能有很多值与中间那个相等，但不一定被分到左边
    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])
            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
            tree[dep+1][rpos++] = tree[dep][i];
        toleft[dep][i] = toleft[dep][l-1] + lpos -l;
    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
 }


 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];  //所查找区间放在左树中的个数
     if(cnt >= k)
     {   //得到l左边放到左子树的个数，加上L即是开始位置
         int lpos = L+toleft[dep][l-1]-toleft[dep][L-1];
         int rpos = lpos+cnt-1;
         return query(L,mid,lpos,rpos,dep+1,k);
     }
     else
     {   //R-r可以得出后面空出了多少位置
         int rpos = r+toleft[dep][R]-toleft[dep][r];
         int lpos = rpos-(r-l-cnt);
         return query(mid+1,R,lpos,rpos,dep+1,k-cnt);
     }
 }


 int main()
 {
     int T,n,m;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         memset(tree,0,sizeof(tree));
         for(int i = 1;i <= n;i++)
         {
             scanf("%d",&sorted[i]);
             tree[0][i] = sorted[i];
         }
         sort(sorted+1,sorted+n+1);
         build(1,n,0);
         int l,r,k;
         while(m--)
         {
             scanf("%d%d%d",&l,&r,&k);
             printf("%d\n",query(1,n,l,r,0,k));
         }
     }
     return 0;
 }