<p>Given an <code>n x n</code> binary matrix <code>grid</code>, return <em>the length of the shortest <strong>clear path</strong> in the matrix</em>. If there is no clear path, return <code>-1</code>.</p>

<p>A <strong>clear path</strong> in a binary matrix is a path from the <strong>top-left</strong> cell (i.e., <code>(0, 0)</code>) to the <strong>bottom-right</strong> cell (i.e., <code>(n - 1, n - 1)</code>) such that:</p>

<ul>
	<li>All the visited cells of the path are <code>0</code>.</li>
	<li>All the adjacent cells of the path are <strong>8-directionally</strong> connected (i.e., they are different and they share an edge or a corner).</li>
</ul>

<p>The <strong>length of a clear path</strong> is the number of visited cells of this path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example1_1.png" style="width: 500px; height: 234px;" />
<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/example2_1.png" style="height: 216px; width: 500px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,0]]
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j] is 0 or 1</code></li>
</ul>

const static int dir[9] = {0, 1, 0, -1, 1, 1, -1, -1, 0};

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        const int n = grid.size();
        if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1) return -1;
        vector<pair<int, int>> v{{0, 0}};
        grid[0][0] = 2;
        int count = 1;
        while (!v.empty() && grid[n - 1][n - 1] == 0) {
            decltype(v) next;
            for (const auto& p: v) {
                int x = p.first, y = p.second;
                for (int i = 0; i < 8; ++i) {
                    int nx = x + dir[i];
                    int ny = y + dir[i+1];
                    if (0 <= nx && nx < n && 0 <= ny && ny < n && grid[nx][ny] == 0) {
                        next.push_back({nx, ny});
                        grid[nx][ny] = 2;
                    }
                }
            }
            v = next;
            ++count;
        }
        if (grid[n - 1][n - 1] == 0) return -1;
        return count;
    }
};
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