<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:
    void dfs(int ind,vector<vector<int>>&v,vector<bool>&visit,vector<int>&v1)
    {
        visit[ind] = 1;
        for(auto k1 : v[ind])
        {
            if(!visit[k1])
            {
                dfs(k1,v,visit,v1);
            }
        }
        v1.push_back(ind);
    }
    const int dx[4] = {1,-1,0,0},dy[4] = {0,0,1,-1};
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int m = matrix[0].size();
        vector<vector<int>>v = matrix;
        vector<vector<int>>v1(n*m);
        for(int i = 0;i<n;i++)
        {
            for(int j= 0;j<m;j++)
            {
                for(int k = 0;k<4;k++)
                {
                    int x1 = dx[k]+i,y1 = dy[k]+j;
                    if(!(x1 == -1 || x1 == n || y1 == -1 || y1 == m))
                    {
                        if(v[x1][y1] > v[i][j]) v1[m*i+j].push_back(x1*m+y1);
                    }
                }
            }
        }   
        vector<int>topo;
        vector<bool>visit(n*m,0);
        for(int i = 0;i<n*m;i++)
        {
            if(!visit[i])
            dfs(i,v1,visit,topo);
        }
        reverse(topo.begin(),topo.end());
        vector<vector<int>>v2(n*m);
        for(int i = 0;i<n*m;i++)
        {
            for(int j = 0;j<v1[i].size();j++)
            {
                v2[v1[i][j]].push_back(i);
            }
        }
        vector<int>dp(n*m,1);
        // return topo.size();
        for(auto k1 : topo)
        {
            // cout<<k1<<endl;
            for(auto k2 : v2[k1])
            {
                dp[k1] = max(dp[k1],dp[k2]+1);
            }
        }
        int ans = 0;
        for(int i = 0;i<n*m;i++)
        {
            ans = max(ans,dp[i]);
        }
        return ans;
    }
};
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