<p>An <code>n x n</code> grid is composed of <code>1 x 1</code> squares where each <code>1 x 1</code> square consists of a <code>&#39;/&#39;</code>, <code>&#39;\&#39;</code>, or blank space <code>&#39; &#39;</code>. These characters divide the square into contiguous regions.</p>

<p>Given the grid <code>grid</code> represented as a string array, return <em>the number of regions</em>.</p>

<p>Note that backslash characters are escaped, so a <code>&#39;\&#39;</code> is represented as <code>&#39;\\&#39;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/1.png" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [&quot; /&quot;,&quot;/ &quot;]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/2.png" style="width: 200px; height: 198px;" />
<pre>
<strong>Input:</strong> grid = [&quot; /&quot;,&quot;  &quot;]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/15/4.png" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [&quot;/\\&quot;,&quot;\\/&quot;]
<strong>Output:</strong> 5
<strong>Explanation: </strong>Recall that because \ characters are escaped, &quot;\\/&quot; refers to \/, and &quot;/\\&quot; refers to /\.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>&#39;/&#39;</code>, <code>&#39;\&#39;</code>, or <code>&#39; &#39;</code>.</li>
</ul>

class Solution {
public:

    void dfs(vector<vector<int>>& square,int n,vector<vector<bool>>& visited,int r,int c)
    {
        visited[r][c]=true;

        int nr,nc;
        for(int i=-1;i<=1;i++)
        {
            nr=r+i;
            for(int j=-1;j<=1;j++)
            {
                nc=c+j;
                if((nr>=0&&nr<n&&nc>=0&&nc<n)&&(nr==r||nc==c)&&(!(nr==r&&nc==c))&&square[nr][nc]==0&&visited[nr][nc]==false)
                {
                    dfs(square,n,visited,nr,nc);
                }
            }
        }
    }


    int regionsBySlashes(vector<string>& grid) 
    {
        int n=grid.size();
        vector<vector<int>> square(n*3,vector<int>(n*3,0));
        for(int i=0;i<n;i++)
        {
            int j=0;
            while(j<n)
            {
                if(grid[i][j]=='\\')
                {
                    square[i*3][j*3]=1;
                    square[i*3+1][j*3+1]=1;
                    square[i*3+2][j*3+2]=1;           
                }
                else if(grid[i][j]=='/')
                {
                    square[i*3][j*3+2]=1;
                    square[i*3+1][j*3+1]=1;
                    square[i*3+2][j*3]=1;
                }
                j++;
            }    
        }

        vector<vector<bool>> visited(n*3,vector<bool>(n*3,false));
        int cnt=0;
        for(int i=0;i<n*3;i++)
        {
            for(int j=0;j<n*3;j++)
            {
                if(visited[i][j]==false&&square[i][j]==0)
                {
                    dfs(square,n*3,visited,i,j);
                    cnt+=1;
                }
            }
        }
        return cnt;
    }
};
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