<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:
    vector<int> delx={1,0,-1,0};
    vector<int> dely={0,1,0,-1};
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        vector<vector<int>> graph(n*m+1,vector<int>());
        vector<int> indegree(n*m+1);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                for(int a=0;a<4;a++)
                {
                    int newx=i+delx[a];
                    int newy=j+dely[a];
                    if(newx>=0 && newx<n && newy>=0 && newy<m && matrix[newx][newy]>matrix[i][j])
                    {
                        graph[i*(m)+j+1].push_back(newx*(m)+newy+1);
                        indegree[newx*(m)+newy+1]++;
                    }
                }
            }
        }
        for(int i=1;i<=n*m;i++)
        {
            if(indegree[i]==0)
            {
                // cout<<i<<endl;
            graph[0].push_back(i);
            indegree[i]++;
            }
        }
        queue<int> q;
        vector<int> distance(n*m+1,0);
        distance[0]=0;
        q.push(0);
        int maxi=0;
        while(!q.empty())
        {
         
            int top=q.front();
            //    cout<<top<<endl;
            q.pop();
            maxi=max(maxi,distance[top]);
            for(auto it:graph[top])
            {
                distance[it]=max(distance[top]+1,distance[it]);
                indegree[it]--;
                if(indegree[it]==0)
                {
                    q.push(it);
                }
            }
        }
        return maxi;
    }
};
36515