Codeforces Round #397 D. Artsem and Saunders（构造）

野性酷女 2022-07-12 08:04 183阅读 0赞

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 ![e8df9cdcdcdd9dc483cfffa96e3dfba73bd4911b.png][], and *h*(*g*(*x*)) = *f*(*x*) for all ![6c71d5f0e27576f1e962d0fbae5dce915abbab38.png][], 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*m* numbers *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

题解：

构造题。

因为：  
g(h(x))=x.  
h(g(x))=f(x).  
所以，h(x)=f(f(x)) && 必须有一个 x= f(x).(观察可以发现)  
从而得到 g(x) =g(f(x)).

所以将这些x全部映射到\[1,m\]中，再映射到f(x)和g(x)中就可以了。

AC代码：

#include<bits/stdc++.h>
    using namespace std;
    /*
    因为： 
    g(h(x))=x
    h(g(x))=f(x)
    所以，h(x)=f(f(x)) && 必须有一个 x= f(x)
    从而得到 g(x) =g(f(x))
    */
    
    int f[100000+7],g[100000+7];
    int h[100000+7];
    int main()
    {
        int n;
        int hn=0;
        cin>>n;
        for(int i=1;i<=n;i++){
        	cin>>f[i];
        	if(f[i]==i){ //x=f(x) 
        		hn++;
        		g[i]=hn;
        		h[g[i]]=i; //h(g(x))=f(x)=f(x);
    		}
    	}
    	//for(int i=1;i<=n;i++)cout<<"g[f[i]]="<<g[f[i]]<<endl;
    	for(int i=1;i<=n;i++){
    		if(!g[f[i]])return 0*puts("-1");
    	}
    	cout<<hn<<endl;      
    	                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       
    	for(int i=1;i<=n;i++)cout<<g[f[i]]<<" ";
    	cout<<endl;
    	
    	for(int i=1;i<=hn;i++)
    	cout<<h[i]<<" ";
    	return 0;
    }

[e8df9cdcdcdd9dc483cfffa96e3dfba73bd4911b.png]: http://codeforces.com/predownloaded/e8/df/e8df9cdcdcdd9dc483cfffa96e3dfba73bd4911b.png
[6c71d5f0e27576f1e962d0fbae5dce915abbab38.png]: /images/20220711/70ccc8fea60e4646b1cd7af8e32c957a.png