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
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) |
Sample Input 2
2
1 2
Sample Output 2
1 2
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
1 # include <cstdio>
2 # include <cstring>
3 # include <cstdlib>
4 # include <iostream>
5 # include <vector>
6 # include <queue>
7 # include <stack>
8 # include <map>
9 # include <set>
10 # include <cmath>
11 # include <algorithm>
12 using namespace std;
13 # define lowbit(x) ((x)&(-x))
14 # define pi acos(-1.0)
15 # define LL long long
16
17 # define eps 1e-8
18 # define MOD 1000000007
19 # define INF 0x3f3f3f3f
20 inline int Scan() {
21 int x=0,f=1; char ch=getchar();
22 while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
23 while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
24 return x*f;
25 }
26 inline void Out(int a) {
27 if(a<0) {putchar('-'); a=-a;}
28 if(a>=10) Out(a/10);
29 putchar(a%10+'0');
30 }
31 const int MX=200005;
32 //Code begin....
33
34 int n;
35 int dat[MX];
36 int ans[MX];
37 int mn[MX];
38 int st[2][MX][40];
39 struct Qu
40 {
41 int l,r;
42 int dex; //zui xiao ,,biao hao
43 bool operator < (const Qu b) const{
44 return dat[dex]>dat[b.dex];
45 }
46 };
47 void Init()
48 {
49 mn[0]=-1;
50 dat[0]=INF;
51 for (int i=1;i<=n;i++)
52 {
53 if ((i&(i-1))==0) mn[i]=mn[i-1]+1;
54 else mn[i]=mn[i-1];
55
56 if (i&1) st[1][i][0]=i;
57 else st[0][i][0]=i;
58 }
59
60 for (int j=1;(1<<j)<=n;j++)
61 {
62 for (int i=1;(i+(1<<j)-1)<=n;i++)
63 {
64 if (dat[st[0][i][j-1]]<=dat[st[0][i+(1<<(j-1))][j-1]])
65 st[0][i][j] = st[0][i][j-1];
66 else
67 st[0][i][j] = st[0][i+(1<<(j-1))][j-1];
68
69 if (dat[st[1][i][j-1]]<=dat[st[1][i+(1<<(j-1))][j-1]])
70 st[1][i][j] = st[1][i][j-1];
71 else
72 st[1][i][j] = st[1][i+(1<<(j-1))][j-1];
73 }
74 }
75 }
76 int st_cal(int l,int r,int same)
77 {
78 int k = mn[r-l+1];
79 int c;
80 if (same) c=l%2;
81 else c=(l+1)%2;
82
83 if (dat[st[c][l][k]]<=dat[st[c][r-(1<<k)+1][k]])
84 return st[c][l][k];
85 else
86 return st[c][r-(1<<k)+1][k];
87 }
88
89 int main()
90 {
91 scanf("%d",&n);
92 for (int i=1;i<=n;i++)
93 dat[i] = Scan();
94 Init();
95 priority_queue<Qu> Q;
96 Q.push((Qu){1,n,st_cal(1,n,1)});
97 int pp=0;
98 for (int i=1;i<=n/2;i++)
99 {
100 Qu now = Q.top();Q.pop();
101 int ml = now.dex;
102 int mr = st_cal(ml+1,now.r,1);
103 ans[pp++]=dat[ml];
104 ans[pp++]=dat[mr];
105 if (ml>now.l)
106 Q.push( (Qu){now.l,ml-1,st_cal(now.l,ml-1,1)} );
107 if (mr-1>ml)
108 Q.push( (Qu){ml+1,mr-1,st_cal(ml+1,mr-1,1)} );
109 if (mr<now.r)
110 Q.push( (Qu){mr+1,now.r,st_cal(mr+1,now.r,1)} );
111 }
112 for (int i=0;i<pp-1;i++)
113 printf("%d ",ans[i]);
114 printf("%d
",ans[pp-1]);
115 return 0;
116 }