<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 {
    vector<int> Father;
public:
    int regionsBySlashes(vector<string>& grid) {
        int N = grid.size();
        Father.resize((N + 1) * (N + 1));
        for(int i = 0; i <= N; i++){
            for(int j = 0; j <= N; j++){
                int id = i * (N + 1) + j;
                Father[id] = id;
                if(i == 0 || i == N || j == 0 || j == N){
                    Father[id] = 0;
                }
            }
        }
        int count = 1;
        for(int i = 0; i < N; i++){
            for(int j = 0; j < N; j++){
                int a, b;
                if(grid[i][j] == ' ')
                    continue;
                if(grid[i][j] == '/'){
                    a = i * (N + 1) + j + 1;
                    b = (i + 1) * (N + 1) + j;
                }else if(grid[i][j] == '\\'){
                    a = i * (N + 1) + j;
                    b = (i + 1) * (N + 1) + j + 1;
                }
                if(FindFather(a) != FindFather(b))
                    Union(a, b);
                else
                    count++;
            }
        }
        return count;
    }
    int FindFather(int x){
        if(Father[x] != x)
            Father[x] = FindFather(Father[x]);
        return Father[x];
    }
    void Union(int x, int y){
        x = Father[x];
        y = Father[y];
        if(x < y)
            Father[y] = x;
        else
            Father[x] = y;
    }
};
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