<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>

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
       int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> dist(n, vector<int>(m, INT_MAX));
        queue<vector<int>> q;
        if (grid[0][0] == 1)
        {
            return -1;
        }
        q.push({0, 0, 0});
        dist[0][0] = 0;
        while (!q.empty())
        {
            vector<int> cur = q.front();
            q.pop();
            vector<pair<int, int>> v = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {1, -1}, {-1, -1}, {-1, 1}};
            for (int i = 0; i < 8; i++)
            {
                int delrow = cur[0] + v[i].first;
                int delcol = cur[1] + v[i].second;
                if (delrow >= 0 && delcol >= 0 && delrow < n && delcol < m && grid[delrow][delcol] == 0 && cur[2] + 1 < dist[delrow][delcol])
                {
                    dist[delrow][delcol] = cur[2] + 1;
                    q.push({delrow, delcol, cur[2] + 1});
                }
            }
        }
        if (dist[n - 1][m - 1] == INT_MAX)
        {
            return -1;
        }
        return dist[n - 1][m - 1]+1;
    }
};
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