Uva11520 - 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

#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 15;
int n;
char mp[MAXN][MAXN];
void change(int &x, int &y)
{
y++;
if(y>n)
{
x++;
y=1;
}
}
bool dfs(int x,int y)
{
if(x==n+1)return true;
if(mp[x][y]!='.')
{
change(x,y);
if(dfs(x,y))return true;
}else{
for(int i='A'; i<='Z'; i++)
{
if(mp[x-1][y]!=i&&mp[x+1][y]!=i&&mp[x][y+1]!=i&&mp[x][y-1]!=i)
{
mp[x][y] = i;
int x1 = x,y1 = y;
change(x1,y1);
if(dfs(x1,y1))return true;
mp[x][y] ='.';
}
}
}
return false;
}
int main()
{
int t, cs;
cin>>t;
for(cs=1; cs<=t; cs++)
{
cin>>n;
memset(mp,0,sizeof (mp));
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
cin>>mp[i][j];
}
}
dfs(1,1);
cout<<"Case "<<cs<<":"<<endl;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
cout<<mp[i][j];
}cout<<endl;
}
}
return 0;
}