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uva 11520 - Fill the Square

2013-09-05 15:18 393 查看
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2515

11520 - Fill the Square

Time limit: 1.000 seconds

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 next n lines output the input matrix with the empty cells filled heeding the rules above.

Sample Input

2

3

...

...

...

3

...

A..

...

Sample Output

Case 1:

ABA

BAB

ABA

Case 2:

BAB

ABA

BAB

AC代码:

// UVa11520 Fill the Square

#include<cstdio>

#include<cstring>

const int maxn = 10 + 5;

char grid[maxn][maxn];

int n;

int main() {

int T;

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] == '.') {//没填过的字母才需要填

for(char ch = 'A'; ch <= 'Z'; ch++) {  //按照字典序依次尝试

bool ok = true;

if(i>0 && grid[i-1][j] == ch) ok = false;  //和上面的字母冲突

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; break; } //没有冲突,填进网格,停止继续尝试

}

}

printf("Case %d:\n", kase);

for(int i = 0; i < n; i++) printf("%s\n", grid[i]);

}

return 0;

}
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