hdu 2665 Kth number（划分树）

Kth number

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6972    Accepted Submission(s): 2227


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

 

划分树班子  求区间第K大的数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 100010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int tree[20][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序好的数
int toleft[20][MAXN];//toleft[p][i]表示第i层从1到i有数分到左边

void build(int l,int r,int dep)
{
if(l==r) return;
int mid=(l+r)/2;
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])
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);
}

//查询区间第K大的数 [L,R]是大区间 [l,r]是要查询的小空间
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)/2;
int cnt=toleft[dep][r]-toleft[dep][l-1];
if(cnt>=k)
{
int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
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()
{
// fread;
int n,m;
int tc;
scanf("%d",&tc);
while(tc--)
{
scanf("%d%d",&n,&m);
MEM(tree,0);
for(int i=1;i<=n;i++)
{
scanf("%d",&tree[0][i]);
sorted[i]=tree[0][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;
}