Description
A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i = 1... n. For example, the square of q = [4, 5, 1, 2, 3] is p = q2 = [2, 3, 4, 5, 1].
This problem is about the inverse operation: given the permutation p you task is to find such permutation q that q2 = p. If there are several such q find any of them.
Input
The first line contains integer n (1 ≤ n ≤ 106) — the number of elements in permutation p.
The second line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of permutation p.
Output
If there is no permutation q such that q2 = p print the number "-1".
If the answer exists print it. The only line should contain n different integers qi (1 ≤ qi ≤ n) — the elements of the permutation q. If there are several solutions print any of them.
Sample Input
Input
42 1 4 3
Output
3 4 2 1
Input
42 1 3 4
Output
-1
Input
52 3 4 5 1
Output
4 5 1 2 3
简单题意
给出一个大小为n的置换群p，求一个置换群q，使得q^2=p
胡说题解
首先我们观察q^2是怎么运算的，置换群可以看成一个一个的环，每个点i向a[i]连一条边就是那个图
q^2其实就是把i的边变成a[a[i]]，也就是在环上走两步，然后q原本的环就变了
1.假设原来是奇数环，那么后来还是一个奇数环，只是顺序变了
2.假设原来是偶数环，那么就会拆成两个大小为一半的环
我们再看p
p上的奇数环可能是原来的奇数环，也有可能是偶数环拆开得来的
p上的偶数环只可能是原来的偶数环拆开得来
对于奇数环我们只要把这个环的后半部分与前半部分（先把这个环断开）交替插入就可以构造出原来的那个奇数环
对于偶数环我们就只能找一个相同大小的偶数环交替插入，即两个相同大小的偶数环合并，如果找不到相同大小的偶数环，那么我们就知道不存在这样的q使得q^2=p
1 #include<cstdio>
2 #include<algorithm>
3 using namespace std;
4
5 const int maxn=1000100;
6
7 int n,tot,a[maxn],b[maxn],s[maxn],l[maxn],cir[maxn];
8 bool flag[maxn];
9
10 bool com(int a,int b){
11 return l[a]<l[b];
12 }
13
14 int main(){
15 scanf("%d",&n);
16 int i,j,k;
17 for(i=1;i<=n;i++)scanf("%d",&a[i]);
18 for(i=1;i<=n;i++)
19 if(!flag[i]){
20 cir[++tot]=i;
21 flag[i]=true;
22 ++l[i];
23 j=a[i];
24 while(!flag[j]){
25 flag[j]=true;
26 ++l[i];
27 j=a[j];
28 }
29 }
30 sort(cir+1,cir+1+tot,com);
31 int x=0;
32 bool f=true;
33 for(i=1;i<=tot;i++)
34 if((l[cir[i]]&1)== 0){
35 if(x==0)x=l[cir[i]];
36 else
37 if(x==l[cir[i]])x=0;
38 else f=false;
39 }
40 if(x!=0)f=false;
41 if(f==false)printf("-1");
42 else{
43 for(i=1;i<=tot;i++){
44 if((l[cir[i]]&1)==1){
45 j=cir[i];
46 k=0;
47 while(flag[j]){
48 s[k]=j;
49 flag[j]=false;
50 k=(k+2)%l[cir[i]];
51 j=a[j];
52 }
53 for(j=0;j<l[cir[i]]-1;j++)b[s[j]]=s[j+1];
54 b[s[l[cir[i]]-1]]=s[0];
55 }
56 else{
57 j=cir[i];
58 k=cir[i+1];
59 while(flag[j]){
60 b[j]=k;
61 b[k]=a[j];
62 flag[j]=false;
63 flag[k]=false;
64 j=a[j];
65 k=a[k];
66 }
67 ++i;
68 }
69 }
70 for(i=1;i<=n;i++)printf("%d ",b[i]);
71 }
72 return 0;
73 }
AC代码