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 10^9 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
题意
静态查询区间第 k 小的数。
思路
【划分树/主席树】 的基本应用，以下附模板。
AC 代码
划分树
#include <iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef __int64 LL;
const int maxn=1e5+10;
int tree[20][maxn]; // 每层每个位置的值
int sorted[maxn]; // 已经排序好的数
int toleft[20][maxn]; // 第 i 层从 1-j 有多少个数被划分到了左边
void build(int l,int r,int dep)
{
if(l==r)return;
int mid = (l+r)>>1;
int same = mid-l+1; // 等于中间值且被划分到左边数的个数
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);
}
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()
{
ios::sync_with_stdio(false);
int n,m;
while(cin>>n>>m)
{
memset(tree,0,sizeof(tree));
for(int i=1; i<=n; i++)
{
cin>>tree[0][i];
sorted[i]=tree[0][i];
}
sort(sorted+1,sorted+n+1);
build(1,n,0);
int s,t,k;
while(m--)
{
cin>>s>>t>>k;
cout<<query(1,n,s,t,0,k)<<endl;
}
}
return 0;
}
主席树
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 2000000;
int a[maxn],b[maxn];
int sum[maxn];
int ls[maxn];
int rs[maxn];
int rk[maxn];
int tot;
void build(int &o,int l,int r)
{
o=++tot;
sum[o]=0;
if(l==r)return;
int mid = (l+r)>>1;
build(ls[o],l,mid);
build(rs[o],mid+1,r);
}
void update(int &o,int l,int r,int last,int p)
{
o = ++tot;
ls[o]=ls[last];
rs[o]=rs[last];
sum[o]=sum[last]+1;
if(l==r)return;
int mid = (l+r)>>1;
if(p<=mid)update(ls[o],l,mid,ls[last],p);
else update(rs[o],mid+1,r,rs[last],p);
}
int query(int L,int R,int l,int r,int k)
{
if(L==R)return L;
int mid = (L+R)>>1;
int cnt = sum[ls[r]]-sum[ls[l]];
if(k<=cnt)return query(L,mid,ls[l],ls[r],k);
else return query(mid+1,R,rs[l],rs[r],k-cnt);
}
int main()
{
ios::sync_with_stdio(false);
int n,m;
while(cin>>n>>m)
{
tot=0;
for(int i=1; i<=n; i++)
cin>>a[i],b[i]=a[i];
sort(b+1,b+n+1);
int sz = unique(b+1,b+n+1)-(b+1);
build(rk[0],1,sz);
for(int i=1; i<=n; i++)
{
a[i] = lower_bound(b+1,b+n+1,a[i])-b;
update(rk[i],1,sz,rk[i-1],a[i]);
}
while(m--)
{
int l,r,k;
cin>>l>>r>>k;
cout<<b[query(1,sz,rk[l-1],rk[r],k)]<<endl;
}
}
return 0;
}
