C. Restoring Permutation

最新推荐文章于 2021-08-25 20:06:09 发布

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最新推荐文章于 2021-08-25 20:06:09 发布

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分类专栏： codeforces ACM 贪心

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本文解析了CodeForces竞赛中一道关于寻找字典序最小排列的问题，输入为一系列整数，目标是找到满足特定条件的最小排列，或判断是否不可能实现。通过示例输入输出，介绍了算法思路和C++实现代码。

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链接：https://codeforces.ml/contest/1315/problem/C

You are given a sequence b1,b2,…,bnb1,b2,…,bn. Find the lexicographically minimal permutation a1,a2,…,a2na1,a2,…,a2n such that bi=min(a2i−1,a2i)bi=min(a2i−1,a2i), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases tt (1≤t≤1001≤t≤100).

The first line of each test case consists of one integer nn — the number of elements in the sequence bb (1≤n≤1001≤n≤100).

The second line of each test case consists of nn different integers b1,…,bnb1,…,bn — elements of the sequence bb (1≤bi≤2n1≤bi≤2n).

It is guaranteed that the sum of nn by all test cases doesn't exceed 100100.

Output

For each test case, if there is no appropriate permutation, print one number −1−1.

Otherwise, print 2n2n integers a1,…,a2na1,…,a2n — required lexicographically minimal permutation of numbers from 11 to 2n2n.

Example

input

Copy

output

Copy

1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,t,r,q,k,s,max1=0,p;
long long a[10001],b[10001];
map<long long ,long long>m; 
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>n;
		m.clear();
		for(int i=1;i<=n;i++)
		{
			cin>>b[i];
			m[b[i]]=1;
		}
		int flag=1;
		for(int i=1;i<=n;i++)
		{
			a[i*2-1]=b[i];
			k=b[i];
			while(m[k]==1)
			{
				k++;
			}
			if(k<=2*n)
			{
				a[i*2]=k;
				m[k]=1;
			}
			else
			{
				flag=0;
				break;
			}
		}
		if(flag==0)
		{
			cout<<-1<<endl;
		}
		else
		{
			for(int i=1;i<=2*n;i++)
			{
				cout<<a[i]<<" ";
			}
			cout<<endl;
		}
		 
	}
}