<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 count, n;
   vector<int> f;
    
   int regionsBySlashes(vector<string>& grid) {
        n = grid.size();
       // a square is splitted into 4 triangles
        count = n * n * 4;
        for (int i = 0; i < count; ++i)
            f.push_back(i);
        //     0
        // 3       1
        //     2
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                // 2: down; 0: up
                if (i > 0) 
                    uni(g(i - 1, j, 2), g(i, j, 0));
                // 1: right; 3: left
                if (j > 0) 
                    uni(g(i , j - 1, 1), g(i , j, 3));
                
                if (grid[i][j] != '/') {
                    uni(g(i , j, 0), g(i , j,  1));
                    uni(g(i , j, 2), g(i , j,  3));
                }
                if (grid[i][j] != '\\') {
                    uni(g(i , j, 0), g(i , j,  3));
                    uni(g(i , j, 2), g(i , j,  1));
                }
            }
        }
        return count;
    }

    int find(int x) {
        if (x != f[x]) {
            f[x] = find(f[x]);
        }
        return f[x];
    }
    
    void uni(int x, int y) {
        x = find(x); y = find(y);
        if (x != y) {
            f[x] = y;
            count--;
        }
    }
    
    int g(int i, int j, int k) {
        return (i * n + j) * 4 + k;
    }
};








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