K-th Number
| Time Limit: 20000MS | Memory Limit: 65536K | |
| Total Submissions: 43315 | Accepted: 14296 | |
| 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.
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).
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
给定一个区间，求区间的第k大小的值
该题求的是静态区间第k大，是主席树的入门题目，但是划分树也可以轻松解决，下面两个代码都有
主席树代码
/* ***********************************************
Author :kuangbin
Created Time :2013-9-4 20:13:20
File Name :POJ2104.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std; const int MAXN = ;
const int M = MAXN * ;
int n,q,m,tot;
int a[MAXN], t[MAXN];
int T[M], lson[M], rson[M], c[M]; void Init_hash()
{
for(int i = ; i <= n;i++)
t[i] = a[i];
sort(t+,t++n);
m = unique(t+,t++n)-t-;
}
int build(int l,int r)
{
int root = tot++;
c[root] = ;
if(l != r)
{
int mid = (l+r)>>;
lson[root] = build(l,mid);
rson[root] = build(mid+,r);
}
return root;
}
int hash(int x)
{
return lower_bound(t+,t++m,x) - t;
}
int update(int root,int pos,int val)
{
int newroot = tot++, tmp = newroot;
c[newroot] = c[root] + val;
int l = , r = m;
while(l < r)
{
int mid = (l+r)>>;
if(pos <= mid)
{
lson[newroot] = tot++; rson[newroot] = rson[root];
newroot = lson[newroot]; root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++; lson[newroot] = lson[root];
newroot = rson[newroot]; root = rson[root];
l = mid+;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int left_root,int right_root,int k)
{
int l = , r = m;
while( l < r)
{
int mid = (l+r)>>;
if(c[lson[left_root]]-c[lson[right_root]] >= k )
{
r = mid;
left_root = lson[left_root];
right_root = lson[right_root];
}
else
{
l = mid + ;
k -= c[lson[left_root]] - c[lson[right_root]];
left_root = rson[left_root];
right_root = rson[right_root];
}
}
return l;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(scanf("%d%d",&n,&q) == )
{
tot = ;
for(int i = ;i <= n;i++)
scanf("%d",&a[i]);
Init_hash();
T[n+] = build(,m);
for(int i = n;i ;i--)
{
int pos = hash(a[i]);
T[i] = update(T[i+],pos,);
}
while(q--)
{
int l,r,k;
scanf("%d%d%d",&l,&r,&k);
printf("%d\n",t[query(T[l],T[r+],k)]);
}
}
return ;
}
划分树代码
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std; const int MAXN=;
int tree[][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
int toleft[][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)>>;
int same=mid-l+;//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+;
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+][lpos++]=tree[dep][i];
else if(tree[dep][i]==sorted[mid]&&same>)
{
tree[dep+][lpos++]=tree[dep][i];
same--;
}
else //比中间值大分入右边
tree[dep+][rpos++]=tree[dep][i];
toleft[dep][i]=toleft[dep][l-]+lpos-l;//从1到i放左边的个数 }
build(l,mid,dep+);
build(mid+,r,dep+); } //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)>>;
int cnt=toleft[dep][r]-toleft[dep][l-];//[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-]-toleft[dep][L-];
//左端点加上查询区间会被放在左边的个数
int newr=newl+cnt-;
return query(L,mid,newl,newr,dep+,k);
}
else
{
int newr=r+toleft[dep][R]-toleft[dep][r];
int newl=newr-(r-l-cnt);
return query(mid+,R,newl,newr,dep+,k-cnt);
}
} int main(){
int T;
int n,m;
int s,t,k;
int cnt=;
while(scanf("%d%d",&n,&m)!=EOF)
{
cnt++;
//scanf("%d%d",&n,&m);
memset(tree,,sizeof(tree));//这个必须
for(int i=;i<=n;i++)//从1开始
{
scanf("%d",&tree[][i]);
sorted[i]=tree[][i];
}
sort(sorted+,sorted+n+);
build(,n,); // printf("Case %d:\n",cnt);
while(m--)
{
scanf("%d%d%d",&s,&t,&k);
printf("%d\n",query(,n,s,t,,k));
}
}
return ;
}