<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>

#define F first
#define S second
#define INF 10001
using state = pair<int,int>;

class Solution {
public:
    vector<vector<int>> dis;
    int n;
    vector<state> neighbours(state node,vector<vector<int>>& grid){
        vector<state> nei;
        int dx[8] = {1,-1,0,0,1,1,-1,-1};
        int dy[8] = {0,0,1,-1,1,-1,1,-1};
        for(int i=0;i<8;i++){
            int x= node.F+dx[i];
            int y = node.S+dy[i];
            if(x>=0&&y>=0&&x<n&&y<n&&grid[x][y]==0){nei.push_back({x,y});}
        }
        return nei;
    }
    void bfs(vector<vector<int>>& grid){
        if(grid[0][0]){return;}
        queue<state> q;
        q.push({0,0});dis[0][0] = 0;
        while(!q.empty()){
            auto node = q.front();q.pop();
            for(auto v : neighbours(node,grid)){
                if(dis[v.F][v.S] ==INF){
                    q.push(v);dis[v.F][v.S]=dis[node.F][node.S]+1;
                }
            }
        }
    }
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        n = grid.size();
        dis.assign(n,vector<int> (n,INF));
        bfs(grid);
        return (dis[n-1][n-1]==INF)? -1: dis[n-1][n-1]+1;
    }
};
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