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。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define MP make_pair 6 #define PB push_back 7 typedef long long LL; 8 typedef pair<int,int> PII; 9 const double eps=1e-8; 10 const double pi=acos(-1.0); 11 const int K=3e5+7; 12 const int mod=1e9+7; 13 14 int n,vb[4*K],vc[4*K],ff[4*K],hs[K],a[K]; 15 //vb奇数，vc偶数 16 void push_down(int o) 17 { 18 if(ff[o]) 19 { 20 swap(vb[o<<1],vc[o<<1]); 21 swap(vb[o<<1|1],vc[o<<1|1]); 22 ff[o<<1]^=ff[o],ff[o<<1|1]^=ff[o]; 23 ff[o]=0; 24 } 25 } 26 void update(int o,int l,int r,int pos,int x) 27 { 28 if(l==r) 29 { 30 if(pos&1) vb[o]=x,vc[o]=K; 31 else vb[o]=K,vc[o]=x; 32 return ; 33 } 34 int mid=l+r>>1; 35 push_down(o); 36 if(pos<=mid) update(o<<1,l,mid,pos,x); 37 else update(o<<1|1,mid+1,r,pos,x); 38 vb[o]=min(vb[o<<1],vb[o<<1|1]); 39 vc[o]=min(vc[o<<1],vc[o<<1|1]); 40 } 41 void update2(int o,int l,int r,int nl,int nr) 42 { 43 if(l==nl&&r==nr) 44 { 45 ff[o]^=1;swap(vb[o],vc[o]); 46 return ; 47 } 48 int mid=l+r>>1; 49 push_down(o); 50 if(nr<=mid) update2(o<<1,l,mid,nl,nr); 51 else if(nl>mid) update2(o<<1|1,mid+1,r,nl,nr); 52 else update2(o<<1,l,mid,nl,mid),update2(o<<1|1,mid+1,r,mid+1,nr); 53 vb[o]=min(vb[o<<1],vb[o<<1|1]); 54 vc[o]=min(vc[o<<1],vc[o<<1|1]); 55 } 56 int query(int o,int l,int r,int nl,int nr) 57 { 58 if(l==nl&&r==nr)return vb[o]; 59 int mid=l+r>>1; 60 push_down(o); 61 if(nr<=mid) return query(o<<1,l,mid,nl,nr); 62 else if(nl>mid) return query(o<<1|1,mid+1,r,nl,nr); 63 else return min(query(o<<1,l,mid,nl,mid),query(o<<1|1,mid+1,r,mid+1,nr)); 64 } 65 struct node 66 { 67 int l,r,pl,pr; 68 node(){} 69 node(int i,int j,int p,int q){l=i,r=j,pl=p,pr=q;} 70 bool operator<(const node &ta)const 71 { 72 if(a[pl]==a[ta.pl]) return a[pr]>a[ta.pr]; 73 return a[pl]>a[ta.pl]; 74 } 75 }; 76 node sc(int l,int r) 77 { 78 int x=query(1,1,n,l,r); 79 update(1,1,n,hs[x],K); 80 if(hs[x]+1<=r) 81 update2(1,1,n,hs[x]+1,r); 82 int y=query(1,1,n,hs[x]+1,r); 83 update(1,1,n,hs[y],K); 84 if(hs[y]+1<=r) 85 update2(1,1,n,hs[y]+1,r); 86 return node(l,r,hs[x],hs[y]); 87 } 88 priority_queue<node>q; 89 int main(void) 90 { 91 scanf("%d",&n); 92 for(int i=1,mx=n*4;i<=mx;i++) vb[i]=vc[i]=K; 93 for(int i=1;i<=n;i++) 94 { 95 scanf("%d",a+i); 96 update(1,1,n,i,a[i]); 97 hs[a[i]]=i; 98 } 99 q.push(sc(1,n)); 100 while(q.size()) 101 { 102 node tmp=q.top(); 103 q.pop(); 104 printf("%d %d ",a[tmp.pl],a[tmp.pr]); 105 if(tmp.pl-1>tmp.l) q.push(sc(tmp.l,tmp.pl-1)); 106 if(tmp.pl+1<tmp.pr-1) q.push(sc(tmp.pl+1,tmp.pr-1)); 107 if(tmp.pr+1<tmp.r) q.push(sc(tmp.pr+1,tmp.r)); 108 } 109 return 0; 110 }