uva 11520 Fill the Square

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:

1. Make sure no adjacent cells contain the same letter; two cells are adjacent if they share a common edge.
2. 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 next n lines output the input matrix with the empty cells filled heeding the rules above.

Sample Input                       Output for Sample Input  

题目大意：给出一张图，‘.‘的地方要用大写字母去填，要求说每个位置的上下左右如果存在的话所填的字母不能和当前位置相同。输出从上到下，从左到右字典序最小的方案。

解题思路：暴力枚举。

#include<stdio.h>
#include<string.h>
char gra[15][15];
int main() {
	int T, Case = 1;
	scanf("%d%*c", &T);
	while (T--) {
		memset(gra, 0, sizeof(gra));
		int n;
		scanf("%d", &n);
		getchar();
		char temp;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				scanf("%c", &gra[i][j]);
			}
			getchar();
		}
		for (int i = 1; i <= n; i++) {
			for (int  j = 1; j <= n; j++) {
				if (gra[i][j] == '.') {
					for (int k = 0; k < 26; k++) { //字典序
						int ch = k + 'A';
						if (ch == gra[i - 1][j]) continue;//冲突检测
						if (ch == gra[i + 1][j]) continue;
						if (ch == gra[i][j - 1]) continue;
						if (ch == gra[i][j + 1]) continue;
						gra[i][j] = ch;
						break;
					}
				}
				else continue;
			}
		}
		printf("Case %d:\n", Case++);
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				printf("%c", gra[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}