Matrix of Differences（构造）

题目链接

题目描述：

For a square matrix of integers of size $n\times n$, let's define its beauty as follows: for each pair of side-adjacent elements $x$ and $y$, write out the number $|x-y|$, and then find the number of different numbers among them.

For example, for the matrix $\left(\begin{array}{cc}1& 4\\ 3& 2\end{array}\right)$ the numbers we consider are $|1-3|=2$, $|1-4|=3$, $|3-2|=1$ and $|4-2|=2$; there are $3$ different numbers among them $(2,3and1)$, which means that its beauty is equal to $3$.

You are given an integer $n$. You have to find a matrix of size $n\times n$, where each integer from $1$ to $n^2$ occurs exactly once, such that its beauty is the maximum possible among all such matrices.

输入描述：

The first line contains a single integer $t(1\le t\le 49)$ – the number of test cases.

The first (and only) line of each test case contains a single integer $n(2\le n\le 50)$.

输出描述：

For each test case, print $n$ rows of $n$ integers — a matrix of integers of size $n\times n$, where each number from $1$ to $n^2$ occurs exactly once, such that its beauty is the maximum possible among all such matrices. If there are multiple answers, print any of them.

样例：

input:

2
2
3

output:

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

AC代码：

#include <bits/stdc++.h>
 
using namespace std;
 
int n;
 
void solve()
{
	cin >> n;
 
	// 相当与一个n * n的二维数组，可以用a[][]来表示，但n不宜太大
	vector<vector<int>> a(n, vector<int> (n));
 
	int l = 1, r = n * n, t = 0;
 
	// 蛇形排列
	// 第一行从左到右，第二行从右到左，第三行从左到右...依此类推
	for(int i = 0; i < n; i ++)
	{
		for(int j = 0; j < n; j ++)
		{
			if(t)
			{
				a[i][j] = l ++;
			}
			else
			{
				a[i][j] = r --;
			}
 
			t ^= 1;
		}
 
		if(i & 1)
		{
			reverse(a[i].begin(), a[i].end());
		}
	}
 
	for(int i = 0; i < n; i ++)
	{
		for(int j = 0; j < n; j ++)
		{
			// 相当于一个字符串a = " \n";
			// 输出a[0] 或者 a[1];
			cout << a[i][j] << " \n"[j == n - 1];
		}
	}
}
 
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 
	int T;
	cin >> T;
 
	while(T --)
		solve();
 
	return 0;
}