【Codeforces 756 D. Artsem and Saunders】+ 思惟 + 构造

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 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

给出数组 a，是否能够构造出数组g,h知足 ：g(h(x)) = x ，h(g(x)) = f(x)
数组h不重复记录数组a里的元素，数组g(x)记录a[x]在h数组里的位置，但若a[x] != x,便没法构成

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
const int K = 1e5 + 10;
int a[K],b[K],c[K],ok[K];
int main()
{
    int N;
    scanf("%d",&N);
    for(int i = 1; i <= N; i++) scanf("%d",&a[i]),ok[a[i]] = 1;
    int nl = 0;
    for(int i = K - 1; i >= 1; i--)
        if(ok[i]){
           c[++nl] = i,b[i] = nl;
           if(a[i] != i){printf("-1\n"); return 0;}
       }
    printf("%d\n",nl);
    for(int i = 1; i <= N; i++) printf("%d ",b[a[i]]);printf("\n");
    for(int i = 1; i <= nl; i++) printf("%d ",c[i]);
    return 0;
}