AtCoder Regular Contest 080　E - Young Maids

地址：http://arc080.contest.atcoder.jp/tasks/arc080_c

题目：

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.

思路：

　　不想自己写了，结合我代码看官方题解吧。

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 typedef pair<int,int> PII;
 const double eps=1e-;
 const double pi=acos(-1.0);
 const int K=3e5+;
 const int mod=1e9+;

 int n,vb[*K],vc[*K],ff[*K],hs[K],a[K];
 //vb奇数，vc偶数
 void push_down(int o)
 {
     if(ff[o])
     {
         swap(vb[o<<],vc[o<<]);
         swap(vb[o<<|],vc[o<<|]);
         ff[o<<]^=ff[o],ff[o<<|]^=ff[o];
         ff[o]=;
     }
 }
 void update(int o,int l,int r,int pos,int x)
 {
     if(l==r)
     {
         if(pos&) vb[o]=x,vc[o]=K;
         else vb[o]=K,vc[o]=x;
         return ;
     }
     int mid=l+r>>;
     push_down(o);
     if(pos<=mid) update(o<<,l,mid,pos,x);
     else update(o<<|,mid+,r,pos,x);
     vb[o]=min(vb[o<<],vb[o<<|]);
     vc[o]=min(vc[o<<],vc[o<<|]);
 }
 void update2(int o,int l,int r,int nl,int nr)
 {
     if(l==nl&&r==nr)
     {
         ff[o]^=;swap(vb[o],vc[o]);
         return ;
     }
     int mid=l+r>>;
     push_down(o);
     if(nr<=mid) update2(o<<,l,mid,nl,nr);
     else if(nl>mid) update2(o<<|,mid+,r,nl,nr);
     else update2(o<<,l,mid,nl,mid),update2(o<<|,mid+,r,mid+,nr);
     vb[o]=min(vb[o<<],vb[o<<|]);
     vc[o]=min(vc[o<<],vc[o<<|]);
 }
 int query(int o,int l,int r,int nl,int nr)
 {
     if(l==nl&&r==nr)return vb[o];
     int mid=l+r>>;
     push_down(o);
     if(nr<=mid) return query(o<<,l,mid,nl,nr);
     else if(nl>mid) return query(o<<|,mid+,r,nl,nr);
     else return min(query(o<<,l,mid,nl,mid),query(o<<|,mid+,r,mid+,nr));
 }
 struct node
 {
     int l,r,pl,pr;
     node(){}
     node(int i,int j,int p,int q){l=i,r=j,pl=p,pr=q;}
     bool operator<(const node &ta)const
     {
         if(a[pl]==a[ta.pl]) return a[pr]>a[ta.pr];
         return a[pl]>a[ta.pl];
     }
 };
 node sc(int l,int r)
 {
     int x=query(,,n,l,r);
     update(,,n,hs[x],K);
     if(hs[x]+<=r)
         update2(,,n,hs[x]+,r);
     int y=query(,,n,hs[x]+,r);
     update(,,n,hs[y],K);
     if(hs[y]+<=r)
         update2(,,n,hs[y]+,r);
     return node(l,r,hs[x],hs[y]);
 }
 priority_queue<node>q;
 int main(void)
 {
     scanf("%d",&n);
     for(int i=,mx=n*;i<=mx;i++) vb[i]=vc[i]=K;
     for(int i=;i<=n;i++)
     {
         scanf("%d",a+i);
         update(,,n,i,a[i]);
         hs[a[i]]=i;
     }
     q.push(sc(,n));
     while(q.size())
     {
         node tmp=q.top();
         q.pop();
         printf("%d %d ",a[tmp.pl],a[tmp.pr]);
         if(tmp.pl->tmp.l) q.push(sc(tmp.l,tmp.pl-));
         if(tmp.pl+<tmp.pr-) q.push(sc(tmp.pl+,tmp.pr-));
         if(tmp.pr+<tmp.r) q.push(sc(tmp.pr+,tmp.r));
     }
     return ;
 }