codeforces #397 D. Artsem and Saunders（构造）

最新推荐文章于 2018-04-10 15:50:24 发布

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最新推荐文章于 2018-04-10 15:50:24 发布

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文章标签：构造 codeforces

本文链接：https://blog.youkuaiyun.com/spark__s/article/details/55188876

本文探讨了一个数学函数构造问题，任务是给定一个从整数集到自身的函数f，找到两个辅助函数g和h，使得g(h(x))=x且h(g(x))=f(x)，并提供了解决方案及代码实现。

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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 $\text{[math]}$ , and h(g(x)) = f(x) for all $\text{[math]}$ , 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

构造，先根据h(g(x))=f(x)。m就是f(x)中不相同的个数，h(x)就是f(x)中不相同的数，g(x)就是f(x)在所有数种排第几。然后根据g(h(x))=x判断这个解对不对即可。

#include<iostream>
using namespace std;
int f[100005];
int h[100005];
int g[100005];
int ff[100005];
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&f[i]);
		}
		int m=0;
		for(int i=1;i<=n;i++)
		{
			if(ff[f[i]]==0)
			{
				m++;
				h[m]=f[i];
				ff[f[i]]=m;				
			}
			g[i]=ff[f[i]];
		}
		int f=1;
		for(int i=1;i<=m;i++)
		{
			if(g[h[i]]!=i)
			{
				f=0;
				break;
			}
		}
		if(f==0)
		{
			puts("-1");
		}
		else
		{
			cout<<m<<endl;
			for(int i=1;i<=n;i++)
			{
				cout<<g[i]<<" ";
			}
			cout<<endl;
			for(int i=1;i<=m;i++)
			{
				cout<<h[i]<<" ";
			}
			cout<<endl;
		}
	}
}