UVa 11520 Fill the Square (貪心&字典序)

11520 - Fill the Square

Time limit: 1.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=456&page=show_problem&problem=2515

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        

2 3 ... ... ... 3 ... A.. ...

Case 1: ABA BAB ABA Case 2: BAB ABA BAB

思路：對每個格子用A~Z去試(可以就break)

完整代碼：

/*0.009s*/

#include<cstdio>
#include<cstring>
const int maxn = 11;

char grid[maxn][maxn];

int main(void)
{
	int T,n;
	scanf("%d", &T);
	for (int kase = 1; kase <= T; kase++)
	{
		scanf("%d", &n);
		for (int i = 0; i < n; i++) 
            scanf("%s", grid[i]);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) 
                if (grid[i][j] == '.')  //沒填過的字母才需要填
				{
					for (char ch = 'A'; ch <= 'Z'; ch++)   	//按照字典序依次嘗試
					{
						bool ok = true;
						if (i > 0 && grid[i - 1][j] == ch) ok = false; 	//和上面的字母沖突
						if (i < n - 1 && grid[i + 1][j] == ch) ok = false;
						if (j > 0 && grid[i][j - 1] == ch) ok = false;
						if (j < n - 1 && grid[i][j + 1] == ch) ok = false;
						if (ok)
						{
							grid[i][j] = ch;  //沒有沖突，填進網格，停止繼續嘗試
							break;
						}
					}
				}
		printf("Case %d:\n", kase);
		for (int i = 0; i < n; i++) 
            printf("%s\n", grid[i]);
	}
	return 0;
}