<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();
        if (grid[0][0] == 1) return -1;

        vector<pair<int,int>> dirs = {
            {1, 0}, {1, 1}, {0, 1}, {-1, 1}, {-1, 0}, {-1, -1}, {0, -1}, {1, -1}
        };

        struct pair_hash {
            inline std::size_t operator()(const std::pair<int,int> & v) const {
                return v.first*31+v.second;
            }
        };

        queue<pair<int, int>> que;
        que.push({0, 0});
        unordered_set<pair<int, int>, pair_hash> visited;
        visited.insert({0, 0});
        int step = 0;

        while (!que.empty()) {
            step++;
            int qsize = que.size();
            while (qsize-- > 0) {
                auto [i, j] = que.front();
                que.pop();
                if (i == n - 1 && j == n - 1) return step;
                for (int d = 0 ;  d < 8; d++) {
                    int ni = i + dirs[d].first;
                    int nj = j + dirs[d].second;
                    if (0 <= ni
                        && ni < n
                        && 0 <= nj
                        && nj < n
                        && grid[ni][nj] == 0
                        && visited.find({ni, nj}) == visited.end()) {
                        que.push({ni, nj});
                        visited.insert({ni, nj});
                        // cout << "push " << ni << " " << nj << endl;
                    }
                }
            }
        }
        return -1;
    }
};
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