uva11520 Fill the Square

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 lled
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 lled 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 ll 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 rst 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, rst output `Case #:' (# replaced by case number) and in the next n lines output the
input matrix with the empty cells lled heeding the rules above.
Sample Input
2
3
...
...
...
3
...
A..
...
Sample Output
Case 1:
ABA
BAB
ABA
Case 2:
BAB
ABA

BAB

问题分析：

又是pdf格式，渣渣画质。刘汝佳大白书的水题。就是一个类似于迷宫的遍历。

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
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]=='.')
        {
            for(char ch='A';ch<='Z';ch++)
            {
                bool ok=true;
                if(i>0&&grid[i-1][j]==ch) ok=false;
                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;
                }
            }
        }
       printf("Case %d:\n",kase);
       for(int i=0;i<n;i++)
       printf("%s\n",grid[i]); 
    }
    return 0;
}