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

        vector<pair<int, int>> dirs{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        function<void(int, int)> dfs = [&] (int r, int c) {
            graph[r][c] = -1;
            for (auto& dir : dirs) {
                int n_r = r + dir.first;
                int n_c = c + dir.second;
                if (n_r >= 0 && n_r < n * 3 && n_c >= 0 && n_c < n * 3 && graph[n_r][n_c] == 0) {
                    dfs(n_r, n_c);
                }
            }
            return ;
        };

        int res = 0;
        for (int r = 0; r < n * 3; ++r) {
            for (int c = 0; c < n * 3; ++c) {
                if (graph[r][c] == 0) {
                    dfs(r, c);
                    ++res;
                }
            }
        }

        return res;
    }
};
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