POJ 题目2140 K-th Number（划分树）

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 42155		Accepted: 13855
Case Time Limit: 2000MS

Description
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in
the array segment.

That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"

For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input
The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).

The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.

The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output
For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

Hint
This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
Source
Northeastern Europe 2004, Northern Subregion

ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int tree[30][100100],sorted[100010],toleft[30][100010];
int cmp(const void *a,const void *b)
{
return *(int *)a-*(int *)b;
}
void build(int l,int r,int dep)
{
if(l==r)
{
return;
}
int mid=(l+r)>>1;
int same=mid-l+1;
int i;
for(i=l;i<=r;i++)
if(tree[dep][i]<sorted[mid])
same--;
int lpos=l;
int rpos=mid+1;
for(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;//1 l傻傻分不清楚
}
build(l,mid,dep+1);//又是1 l傻傻分不清楚
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];
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()
{
int n,m;
//	while(scanf("%d%d",&n,&m)!=EOF)
//	{
int i;
scanf("%d%d",&n,&m);
memset(tree,0,sizeof(tree));
for(i=1;i<=n;i++)
{
scanf("%d",&tree[0][i]);
sorted[i]=tree[0][i];
}
qsort(sorted+1,n,sizeof(sorted[1]),cmp);
build(1,n,0);
while(m--)
{
int a,b,k;
scanf("%d%d%d",&a,&b,&k);
int ans=query(1,n,a,b,0,k);
printf("%d\n",ans);
}
//	}
}