<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]||grid[n-1][n-1]) return -1;
        vector<vector<int>>dist(n,vector<int>(n,INT_MAX));
        //dist,node
        int drow[]={-1,-1,-1,0,1,1,1,0};
        int dcol[]={-1,0,1,1,1,0,-1,-1};
        priority_queue<pair<pair<int,int>,int>,vector<pair<pair<int,int>,int>>,greater<pair<pair<int,int>,int>>>pq;
        pq.push({{0,0},1});
        dist[0][0]=1;
        while(!pq.empty()){
            auto it=pq.top();
            pq.pop();
            int r=it.first.first;
            int c=it.first.second;
            int dis=it.second;
            for(int i=0;i<8;i++){
                int nr=r+drow[i];
                int nc=c+dcol[i];
                if(nr>=0&&nr<n&&nc>=0&&nc<n&&dist[nr][nc]>dis+1&&grid[nr][nc]==0){
                    dist[nr][nc]=dis+1;
                    pq.push({{nr,nc},dist[nr][nc]});
                }
            }
        }
      return dist[n-1][n-1]!=INT_MAX?dist[n-1][n-1]:-1;
    }
};
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