<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) {
        int n=matrix.size(),m=matrix[0].size();
        vector<vector<int>> dp(n,vector<int>(m,1));
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            pq.push({matrix[i][j],i,j});
        }
        while(pq.size())
        {
            int ele=pq.top()[0];
            int i=pq.top()[1];
            int j=pq.top()[2];
            pq.pop();
            if(i-1>=0&&matrix[i-1][j]>matrix[i][j])
            dp[i-1][j]=max(dp[i-1][j],dp[i][j]+1);
            if(j-1>=0&&matrix[i][j-1]>matrix[i][j])
            dp[i][j-1]=max(dp[i][j-1],dp[i][j]+1);
            if(i+1<n&&matrix[i+1][j]>matrix[i][j])
            dp[i+1][j]=max(dp[i+1][j],dp[i][j]+1);
            if(j+1<m&&matrix[i][j+1]>matrix[i][j])
            dp[i][j+1]=max(dp[i][j+1],dp[i][j]+1);
        }
        int ans=0;
        for(auto it:dp)
        {
            for(auto itt:it)
            {
                // cout<<itt<<" ";
                ans=max(ans,itt);
            }
            // cout<<endl;
        }
        return ans;
    }
};
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