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

D. Artsem and Saunders

题目连接：

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

Description

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 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.

Sample Input

3
1 2 3

Sample Output

3
1 2 3
1 2 3

Hint

题意

给你n个数f(i)

现在让你构造一个长度为n的g(i)，和一个长度为m的h(i),m由你自己定。

要求g(h(x))=x,h(g(x))=f(x)

题解：

g(h(x))= x

h(g(x)) = f(x)

得h(x) = f(h(x))

继而得到f(x) = f(f(x))

从而能够判断是否能够构造成功

然后h(x)就是f(x)的去重排序

从而得到g

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+7;
int a[maxn],n,b[maxn],c[maxn],m;
map<int,int> H;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        if(a[a[i]]!=a[i]){
            cout<<"-1"<<endl;
            return 0;
        }
    for(int i=1;i<=n;i++){
        if(!H[a[i]]){
            m++;
            H[a[i]]=m;
            c[m]=a[i];
        }
        b[i]=H[a[i]];
    }
    cout<<m<<endl;
    for(int i=1;i<=n;i++)
        cout<<b[i]<<" ";
    cout<<endl;
    for(int i=1;i<=m;i++)
        cout<<c[i]<<" ";
    cout<<endl;
}