Young Maids

E - Young Maids

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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).

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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.

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Sample Input 1

4
3 2 4 1

Sample Output 1

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)

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Sample Input 2

2
1 2

Sample Output 2

1 2

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Sample Input 3

8
4 6 3 2 8 5 7 1

Sample Output 3

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)

//题意：给出一个 p 序列，一个 q 序列，p 中有 1 -- n 的某种排列，q初始为空，现要求可以每次选两个相邻的数，移动到 q 中，要操作到 p 为空，并且要求 q 序列字典序最小

题解：相当难想到的一个题，假如有一个区间 l,r，肯定是，选择这区间内与 l 奇偶相同的，并且最小的作为开头，这样才能保证字典序最小，然后第二个数，应该为选的数右边的，并且与 l 奇偶性不同的，作为第二个，也要保证字典序最小，这部分是标准RMQ问题，随便用什么。然后，区间被分割成最多三块，用优先队列保存区间即可，排队顺序为区间内与左端点奇偶相同的位置上值小的先出队  n*lgn

还是难以想到啊，很好的题

 # include <cstdio>
 # include <cstring>
 # include <cstdlib>
 # include <iostream>
 # include <vector>
 # include <queue>
 # include <stack>
 # include <map>
 # include <set>
 # include <cmath>
 # include <algorithm>
 using namespace std;
 # define lowbit(x) ((x)&(-x))
 # define pi acos(-1.0)
 # define LL long long

 # define eps 1e-
 # define MOD
 # define INF 0x3f3f3f3f
 inline int Scan() {
     int x=,f=; char ch=getchar();
     while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 inline void Out(int a) {
     if(a<) {putchar('-'); a=-a;}
     if(a>=) Out(a/);
     putchar(a%+'');
 }
 const int MX=;
 //Code begin....

 int n;
 int dat[MX];
 int ans[MX];
 int mn[MX];
 int st[][MX][];
 struct Qu
 {
     int l,r;
     int dex; //zui xiao ,,biao hao
     bool operator < (const Qu b) const{
         return dat[dex]>dat[b.dex];
     }
 };
 void Init()
 {
     mn[]=-;
     dat[]=INF;
     for (int i=;i<=n;i++)
     {
         if ((i&(i-))==) mn[i]=mn[i-]+;
         else mn[i]=mn[i-];

         if (i&) st[][i][]=i;
         else st[][i][]=i;
     }

     for (int j=;(<<j)<=n;j++)
     {
         for (int i=;(i+(<<j)-)<=n;i++)
         {
             if (dat[st[][i][j-]]<=dat[st[][i+(<<(j-))][j-]])
                 st[][i][j] = st[][i][j-];
             else
                 st[][i][j] = st[][i+(<<(j-))][j-];

             if (dat[st[][i][j-]]<=dat[st[][i+(<<(j-))][j-]])
                 st[][i][j] = st[][i][j-];
             else
                 st[][i][j] = st[][i+(<<(j-))][j-];
         }
     }
 }
 int st_cal(int l,int r,int same)
 {
     int k = mn[r-l+];
     int c;
     if (same) c=l%;
     else c=(l+)%;

     if (dat[st[c][l][k]]<=dat[st[c][r-(<<k)+][k]])
         return st[c][l][k];
     else
         return st[c][r-(<<k)+][k];
 }

 int main()
 {
     scanf("%d",&n);
     for (int i=;i<=n;i++)
         dat[i] = Scan();
     Init();
     priority_queue<Qu> Q;
     Q.push((Qu){,n,st_cal(,n,)});
     int pp=;
     for (int i=;i<=n/;i++)
     {
         Qu now = Q.top();Q.pop();
         int ml = now.dex;
         int mr = st_cal(ml+,now.r,);
         ans[pp++]=dat[ml];
         ans[pp++]=dat[mr];
         if (ml>now.l)
             Q.push( (Qu){now.l,ml-,st_cal(now.l,ml-,)}  );
         if (mr->ml)
             Q.push( (Qu){ml+,mr-,st_cal(ml+,mr-,)}  );
         if (mr<now.r)
             Q.push( (Qu){mr+,now.r,st_cal(mr+,now.r,)}  );
     }
     for (int i=;i<pp-;i++)
         printf("%d ",ans[i]);
     printf("%d\n",ans[pp-]);
     return ;
 }