填充正方形（Fill the Square）

最新推荐文章于 2023-04-18 12:51:05 发布

Sharing_Li

最新推荐文章于 2023-04-18 12:51:05 发布

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该问题要求在给定的正方形网格中填充空格，使其每个相邻单元格不含有相同字母，同时确保填充的字母顺序是字典序最小。输入包含测试用例数和n×n的字符矩阵，矩阵中'.'表示空格，已有的字母范围为[A-Z]。解题思路是按字典序从'A'开始尝试填充每个空格，直至满足条件。输出应展示填充后的网格，每个案例前标明案号。

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Fill the Square Input: Standard Input

Output: Standard Output

In this problem, you have to draw a square using uppercase English Alphabets.

To be more precise, you will be given a square grid with some empty blocks and others already filled for you with some letters to make your task easier. You have to insert characters in every empty cell so that the whole grid is filled with alphabets. In doing so you have to meet the following rules:

Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common edge.
There could be many ways to fill the grid. You have to ensure you make the lexicographically smallest one. Here, two grids are checked in row major order when comparing lexicographically.

Input The first line of input will contain an integer that will determine the number of test cases. Each case starts with an integer n( n<=10 ), that represents the dimension of the grid. The next n lines will contain n characters each. Every cell of the grid is either a ‘.’ or a letter from [A, Z]. Here a ‘.’ Represents an empty cell.

Output For each case, first output Case #: ( # replaced by case number ) and in the nextnlines output the input matrix with the empty cells filled heeding the rules above.

Sample Input Output for Sample Input

...

A..

...

Case 1:

ABA

BAB

ABA

Case 2:

BAB

ABA

BAB

思路：对每一个‘.’进行从‘A’到‘Z’的填放，直到满足不冲突条件

代码如下：

#include <cstdio>
const int maxn = 12;
char matrix[maxn][maxn];
int n;
int main()
{
	int T;
	scanf("%d",&T);
	for (int m = 1; m <= T; m++)
	{
		scanf("%d",&n);
		for (int i = 0; i < n; i++)
			scanf("%s",matrix[i]);
		for (int j = 0; j < n; j++)
			for (int k = 0; k < n; k++)
				if (matrix[j][k] == '.')
				{
					for (char ch = 'A'; ch <= 'Z'; ch++)//比较填充
					{
						int flag = 1;
						if ( j > 0 && matrix[j- 1][k] == ch || 
                                        j < n - 1 && matrix[j + 1][k] == ch ||
						  k > 0 && matrix[j][k - 1] == ch || 
                                        k < n - 1 && matrix[j][k + 1] == ch)
							flag = 0;
						if (flag)
						{
							matrix[j][k] = ch;
							break;
						}
					}
				}
		printf("Case %d:\n",m);
		for (int k = 0; k < n; k++)
			printf("%s\n",matrix[k]);
	}
	return 0;
}