959. Regions Cut By Slashes

In a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /, \, or blank space. These characters divide the square into contiguous regions.

(Note that backslash characters are escaped, so a \ is represented as "\\".)

Return the number of regions.

Example 1:

Input:
[
  " /",
  "/ "
]
Output: 2
Explanation: The 2x2 grid is as follows:

Example 2:

Input:
[
  " /",
  "  "
]
Output: 1
Explanation: The 2x2 grid is as follows:

Example 3:

Input:
[
  "\\/",
  "/\\"
]
Output: 4
Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.)
The 2x2 grid is as follows:

Example 4:

Input:
[
  "/\\",
  "\\/"
]
Output: 5
Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.)
The 2x2 grid is as follows:

Example 5:

Input:
[
  "//",
  "/ "
]
Output: 3
Explanation: The 2x2 grid is as follows:

Note:

1. 1 <= grid.length == grid[0].length <= 30
2. grid[i][j] is either '/', '\', or ' '.

---

题目：在由 1 x 1 方格组成的 N x N 网格 grid 中，每个 1 x 1 方块由 /、\ 或空格构成。这些字符会将方块划分为一些共边的区域。（请注意，反斜杠字符是转义的，因此 \ 用 “\” 表示。）。返回区域的数目。

思路：参考大神votrubac的解法。将输入的string表达式所代表的网格扩大三倍，如下图（图片源自votrubac，侵删）。其中右下图为扩大后的图，黑色部分表示原始的/或者\。对扩大后的图进行dfs得到最终的结果。

工程代码下载

class Solution {
public:
    int regionsBySlashes(vector<string>& grid) {
        int n = grid.size();
        vector<vector<int>> g(n*3, vector<int>(n*3));
        
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                if(grid[i][j] == '/')
                    g[i*3][j*3+2] = g[i*3+1][j*3+1] = g[i*3+2][j*3] = 1;
                else if(grid[i][j] == '\\')
                    g[i*3][j*3] = g[i*3+1][j*3+1] = g[i*3+2][j*3+2] = 1;
            }
        }
        
        int res = 0;
        for(int i = 0; i < n*3; ++i){
            for(int j = 0; j < n*3; ++j){
                if(g[i][j] == 0){
                    dfs(g, i, j);
                    res += 1;
                }
            }
        }

        return res;
    }
private:
    void dfs(vector<vector<int>>& g, int i, int j){
        if(i < 0 || i >= g.size() || j < 0 || j >= g.size() || g[i][j] == 1)
            return;
        g[i][j] = 1;
        
        dfs(g, i+1, j);        
        dfs(g, i-1, j);
        dfs(g, i, j+1);
        dfs(g, i, j-1);
    }
};