<p>Given an <code>m x n</code> integers <code>matrix</code>, return <em>the length of the longest increasing path in </em><code>matrix</code>.</p>

<p>From each cell, you can either move in four directions: left, right, up, or down. You <strong>may not</strong> move <strong>diagonally</strong> or move <strong>outside the boundary</strong> (i.e., wrap-around is not allowed).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/05/grid1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[9,9,4],[6,6,8],[2,1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest increasing path is <code>[1, 2, 6, 9]</code>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/27/tmp-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> matrix = [[3,4,5],[3,2,6],[2,2,1]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>The longest increasing path is <code>[3, 4, 5, 6]</code>. Moving diagonally is not allowed.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[1]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        priority_queue<vector<int> > pq;
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                pq.push({-1*matrix[i][j],i,j});
            }
        }
        vector<vector<int>> dir = {{0,-1},{-1,0},{0,1},{1,0}};
        vector<vector<int>> dp(matrix.size(),vector<int>(matrix[0].size(),1));

        while(!pq.empty()){
            auto v=pq.top();
            pq.pop();
            int r=v[1],c=v[2];

            for(auto d: dir){
                int r_ = r+d[0],c_ = c+d[1];
                if(r_>=0 && c_>=0 && r_<matrix.size() && c_<matrix[0].size() &&
                      matrix[r_][c_] > -1*v[0]){
                    dp[r_][c_] = max(dp[r_][c_], dp[r][c]+1);
                }
            }
        }

        int ans = 1;
        for(int i=0;i<matrix.size();i++){
            for(int j=0;j<matrix[0].size();j++){
                ans = max(ans,dp[i][j]);
            }
        }
        return ans;
    }
};
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