<p>There is a strange printer with the following two special requirements:</p>

<ul>
	<li>On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle.</li>
	<li>Once the printer has used a color for the above operation, <strong>the same color cannot be used again</strong>.</li>
</ul>

<p>You are given a <code>m x n</code> matrix <code>targetGrid</code>, where <code>targetGrid[row][col]</code> is the color in the position <code>(row, col)</code> of the grid.</p>

<p>Return <code>true</code><em> if it is possible to print the matrix </em><code>targetGrid</code><em>,</em><em> otherwise, return </em><code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/print1.jpg" style="width: 600px; height: 175px;" />
<pre>
<strong>Input:</strong> targetGrid = [[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/print2.jpg" style="width: 600px; height: 367px;" />
<pre>
<strong>Input:</strong> targetGrid = [[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> targetGrid = [[1,2,1],[2,1,2],[1,2,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to form targetGrid because it is not allowed to print the same color in different turns.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == targetGrid.length</code></li>
	<li><code>n == targetGrid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 60</code></li>
	<li><code>1 &lt;= targetGrid[row][col] &lt;= 60</code></li>
</ul>

class Solution {
public:
    int f(int node,vector<int>&vis,vector<int>&path_vis,vector<vector<int>>&v){
        vis[node]=1;
        path_vis[node]=1;

        int fl = 0; 
        for(auto &it:v[node]){
            if(!vis[it]){
                fl |= f(it,vis,path_vis,v);
            }
            else if(path_vis[it]){
                return 1;
            }
        }
        path_vis[node]=0;
        return fl;
    }
    bool isPrintable(vector<vector<int>>& g) {
        int n = g.size();
        int m = g.front().size();
        
        vector<vector<int>>v(61);
        for(int col = 1; col<=60; col++){
            int l = 1e9; int r = -1e9; int b = -1e9; int t = 1e9;
            for(int i = 0; i<n; i++){
                for(int j = 0; j<m; j++){
                    if(g[i][j]==col){
                        l = min(l,j);
                        r = max(r,j);
                        b = max(b,i);
                        t = min(t,i);
                    }
                }
            }
            if(l==1e9) continue;

            for(int i = t; i<=b; i++){
                for(int j = l; j<=r; j++){
                    if(g[i][j]!=col){
                        v[col].push_back(g[i][j]);
                    }
                }
            }
        }
        vector<int>vis(61,0),path_vis(61,0);
        for(int i=1;i<=60;i++){
            if(!vis[i] && f(i,vis,path_vis,v)) return false;
        }
        return true;
    }
};
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