AtCoder Regular Contest 080 E - Young Maids（线段树+优先队列）

E - Young Maids

Time limit : 2sec / Memory limit : 256MB

Score : 800 points

Problem Statement

Let N be a positive even
number.
We have a permutation of (1,2,…,N), p=(p1,p2,…,pN).
Snuke is constructing another permutation of (1,2,…,N), q,
following the procedure below.
First, let q be an empty
sequence. Then, perform the following operation until p becomes empty:
Select two adjacent elements in p, and call them x and y in
order. Remove x and y from p (reducing
the length of p by 2),
and insert x and y,
preserving the original order, at the beginning of q.
When p becomes empty, q will
be a permutation of (1,2,…,N).
Find the lexicographically smallest permutation that can be obtained as q.

Constraints

N is an even number.
2≤N≤2×105
p is a permutation of (1,2,…,N).

Input

Input is given from Standard Input in the following format:

N
p1 p2 … pN

Output

Print the lexicographically smallest permutation, with spaces in between.

Sample Input 1

Copy

4
3 2 4 1

Sample Output 1

Copy

3 1 2 4

The solution above is obtained as follows:

p	q
(3,2,4,1)	()
↓	↓
(3,1)	(2,4)
↓	↓
()	(3,1,2,4)

Sample Input 2

Copy

2
1 2

Sample Output 2

Copy

1 2

Sample Input 3

Copy

8
4 6 3 2 8 5 7 1

Sample Output 3

Copy

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

p	q
(4,6,3,2,8,5,7,1)	()
↓	↓
(4,6,3,2,7,1)	(8,5)
↓	↓
(3,2,7,1)	(4,6,8,5)
↓	↓
(3,1)	(2,7,4,6,8,5)
↓	↓
()	(3,1,2,7,4,6,8,5)

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e6+10;
int a
, pos
, sum1[N<<2], sum2[N<<2], n, ans;
// 1 奇数 , 2 偶数
typedef pair<int,int> pi;
struct node
{
int first,second,l,r;
bool operator <(const node &A)const
{
return A.first<first;
}
};
vector<node>p;
void build(int l,int r,int rt)
{
if(l==r)
{
scanf("%d", &a[l]);
if(l&1) sum1[rt]=a[l],sum2[rt]=0x3f3f3f3f;
else sum1[rt]=0x3f3f3f3f,sum2[rt]=a[l];
pos[a[l]]=l;
return ;
}
int mid=(l+r)/2;
build(l,mid,rt*2);
build(mid+1,r,rt*2+1);
sum1[rt]=min(sum1[rt<<1],sum1[rt<<1|1]);
sum2[rt]=min(sum2[rt<<1],sum2[rt<<1|1]);
return ;
}
int query1(int l,int r,int rt,int L,int R)
{
if(L<=l&&R>=r) return sum1[rt];
int mid=(l+r)/2;
int ans=0x3f3f3f3f;
if(L<=mid) ans=min(ans,query1(l,mid,rt*2,L,R));
if(R>mid) ans=min(ans,query1(mid+1,r,rt*2+1,L,R));
return ans;
}
int query2(int l,int r,int rt,int L,int R)
{
if(L<=l&&R>=r) return sum2[rt];
int mid=(l+r)/2;
int ans=N+1;
if(L<=mid) ans=min(ans,query2(l,mid,rt*2,L,R));
if(R>mid) ans=min(ans,query2(mid+1,r,rt*2+1,L,R));
return ans;
}
priority_queue<node>q;

int main()
{
scanf("%d", &n);
build(1,n,1);
int x=query1(1,n,1,1,n);
int y=query2(1,n,1,pos[x]+1,n);
q.push((node){x,y,1,n});
for(int i=1;i<=n/2;i++)
{
node tmp=q.top();q.pop();
p.push_back(node{tmp.first,tmp.second,0,0});
int al=pos[tmp.first], ar=pos[tmp.second];
if(tmp.l<al-1)
{
if(tmp.l&1)
{
x=query1(1,n,1,tmp.l,al-1);
y=query2(1,n,1,pos[x]+1,al-1);
}
else
{
x=query2(1,n,1,tmp.l,al-1);
y=query1(1,n,1,pos[x]+1,al-1);
}
q.push(node{x,y,tmp.l,al-1});
}
if(al<ar-1)
{
if((al+1)&1)
{
x=query1(1,n,1,al+1,ar-1);
y=query2(1,n,1,pos[x]+1,ar-1);
}
else
{
x=query2(1,n,1,al+1,ar-1);
y=query1(1,n,1,pos[x]+1,ar-1);
}
q.push(node{x,y,al+1,ar-1});
}
if(ar<tmp.r-1)
{
if((ar+1)&1)
{
x=query1(1,n,1,ar+1,tmp.r);
y=query2(1,n,1,pos[x]+1,tmp.r);
}
else
{
x=query2(1,n,1,ar+1,tmp.r);
y=query1(1,n,1,pos[x]+1,tmp.r);
}
q.push(node{x,y,ar+1,tmp.r});
}
}
for(int i=0; i<p.size(); i++)
{
if(i==p.size()-1)printf("%d %d\n",p[i].first,p[i].second);
else printf("%d %d ",p[i].first,p[i].second);
}
return 0;
}