<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>>& a) {
        int n = a.size();
        int m = a[0].size();
        vector<vector<int>> graph(n * m);
        vector<vector<int>> par(n * m);
        vector<int> indegree(n * m , 0);
        for(int i = 0; i<n; i++){
            for(int j = 0; j<m; j++){
                if(i - 1 >= 0 && a[i - 1][j] > a[i][j]) {
                    graph[i*m + j].push_back(i*m + j - m);
                    par[i*m + j - m].push_back(i*m + j);
                    indegree[i*m + j - m]++;
                }
                if(i + 1 < n && a[i + 1][j] > a[i][j]){
                    graph[i*m + j].push_back(i*m + j + m);
                    par[i*m + j + m].push_back(i*m + j);
                    indegree[i*m + j + m]++;
                }
                if(j - 1 >= 0 && a[i][j - 1] > a[i][j]){
                    graph[i*m + j].push_back(i*m + j - 1);
                    par[i*m + j - 1].push_back(i*m + j);
                    indegree[i*m + j - 1]++;
                }
                if(j + 1 < m && a[i][j + 1] > a[i][j]){
                    graph[i*m + j].push_back(i*m + j + 1);
                    par[i*m + j + 1].push_back(i*m + j);
                    indegree[i*m + j + 1]++;
                }
            }
        }
        queue<int> q;
        for(int i = 0; i<n*m; i++) if(indegree[i] == 0) q.push(i);
        vector<int> topo;
        while(!q.empty()){
            int top = q.front();
            q.pop();
            topo.push_back(top);
            for(auto child : graph[top]){
                indegree[child]--;
                if(indegree[child] == 0) q.push(child);
            }
        }
        vector<int> result(m*n , 0);
        int maxi = 0; 
        for(auto i : topo){
            result[i] = 1;
            for(auto parent : par[i]){
                result[i] = max(result[i] , 1 + result[parent]);
            }
            maxi = max(maxi , result[i]);
        }
        return maxi;
    }
};
52759