Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) D. Artsem and Saunders

地址：http://codeforces.com/contest/765/problem/D

题目：

D. Artsem and Saunders

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.

Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.

Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all , and h(g(x)) = f(x) for all , or determine that finding these is impossible.

Input

The first line contains an integer n (1 ≤ n ≤ 105).

The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n).

Output

If there is no answer, print one integer -1.

Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print mnumbers h(1), ..., h(m).

If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.

Examples

input

3
1 2 3

output

3
1 2 3
1 2 3

input

3
2 2 2

output

1
1 1 1
2

input

2
2 1

output

-1

思路：这题思路清晰的人应该可以很快做出来的。

　　g(h(x)) = x ，可以推出m<=n;

　　h(g(x))=f(x),可以推出m>=f(x)中不同元素的个数。

　　先保证第二个条件h(g(x))=f(x)：

　　　　h[]=去重后的f[]。同时记录f[i]在h[]中的hash值：g[i]hash[f[i]]。

　　然后扫一遍g[h[i]]，判断值是否为x。

　　我不知道怎么证明可以这么做，感觉可行后莽一发就ac了。

　　有谁知道了可以交流下。

 #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=1e5+;
 const int mod=1e9+;

 int n,m,ans;
 int h[K],g[K],f[K];
 map<int,int>hs;

 int main(void)
 {
     cin>>n;
     for(int i=,x;i<=n;i++)
     {
         scanf("%d",&x),f[i]=x;
         if(!hs[x])  h[++m]=x,hs[x]=m;
     }
     for(int i=;i<=n;i++)
         g[i]=hs[f[i]];
     for(int i=;i<=m;i++)
     if(g[h[i]]!=i) ans=;
     if(ans)
         printf("-1\n");
     else
     {
         printf("%d\n",m);
         for(int i=;i<=n;i++)
             printf("%d ",g[i]);
         puts("");
         for(int i=;i<=m;i++)
             printf("%d ",h[i]);
         puts("");
     }
     return ;
 }