codeforces 397div2 D Artsem and Saunders

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

题意 找出一个数m，表示有m个数满足 g(h(x)) = x for all , and h(g(x)) = f(x) 同时在第一个条件中 x满足 x属于m。所以g(x)的整个范围是在m里面的。而且满足第二个条件 h的定义域范围是m 值域范围是 值
因为是伪证,所以我们从大范围开始构造起。 再用小范围进行验证。
h里面是 g g的值范围是m，而h的值范围是f[i]，在第二个条件里面 g的i范围是n。h 和 g构造完成 是根据第二个条件构造的

#include <bits/stdc++.h>using namespace std;int f[200010];int ff[200010];int h[200010];int g[200010];int main(){    int n;    while(cin>>n)    {        memset(f,0,sizeof(f));        memset(ff,0,sizeof(ff));        memset(g,0,sizeof(g));        memset(h,0,sizeof(h));        for(int i=1;i<=n;i++)             cin>>f[i];        int m=0;        for(int i=1;i<=n;i++) //因为是伪证,所以我们从大范围开始构造起。        {           //再用小范围进行验证。            if(ff[f[i]]==0)            {                m++;                h[m]=f[i]; //h里面是 g g的值范围是m，而h的值范围是f[i]                ff[f[i]]=m; //记录f[i]的m.            }            g[i]=ff[f[i]]; //在第二个条件里面 g的i范围是n。        }        //h 和 g构造完成 是根据第二个条件构造的        for(int i=1;i<=m;i++)        {            if(g[h[i]]!=i) //根据第一个条件验证            {                printf("-1\n");                return 0;            }        }        printf("%d\n",m );        for(int i=1;i<=n;i++)        {            printf("%d ",g[i] );        }        printf("\n");        for(int i=1;i<=m;i++)            printf("%d ",h[i] );        printf("\n");    }}