<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 dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
    int dy[8] = {0, -1, 0, 1, -1, 1, -1, 1};
    int n,m,ans=1;
    queue<pair<pair<int,int>,int>>q;
    map<pair<int,int>,vector<pair<int,int>>>mp;
    map<pair<int,int>,int>in;
    void topo(){
        while (!q.empty()){
            pair<int,int>nod=q.front().first;
            ans=max(q.front().second,ans);
            int len=q.front().second;
            q.pop();
            for (auto it:mp[nod]) {
               in[it]--;
               if(in[it]==0){
                   q.push({it,len+1});
               }
            }
        }
    }
    bool can(vector<vector<int>>& mat,int i,int j,int p1,int p2){
        if(i>=0&&j>=0&&i<n&&j<m&&mat[p1][p2]<mat[i][j])return true;
        return false;
    }
    int longestIncreasingPath(vector<vector<int>>& mat) {
        n=mat.size(),m=mat[0].size();
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                for (int k = 0; k < 4; ++k) {
                    int x=dx[k]+i,y=dy[k]+j;
                    if(x>=0&&y>=0&&x<n&&y<m){
                        if(can(mat,x,y,i,j)){
                            mp[{i,j}].push_back({x,y});
                             in[{x,y}]++;
                        }
                    }
                }
            }
        }
        for (auto it:mp) {
           if(in[it.first]==0){
               q.push({it.first,1});
           }
        }
        topo();
        return ans;
    }
};
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