<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:
    bool isPrintable(vector<vector<int>>& targetGrid) {
        vector<int> left(61, INT_MAX);
        vector<int> right(61, INT_MIN);
        vector<int> top(61, INT_MAX);
        vector<int> down(61, INT_MIN);

        map<int, set<int>> next;
        map<int, int> inDegree;

        int n=targetGrid.size(), m=targetGrid[0].size();
        set<int> countAll;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int color = targetGrid[i][j];
                inDegree[color] = 0;
                countAll.insert(color);
                left[color] = min(left[color], j);
                right[color] = max(right[color], j);
                top[color] = min(top[color], i);
                down[color] = max(down[color], i);
            }
        }

        

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int targetColor = targetGrid[i][j];
                for(int color=1;color<=60;color++){
                    if(color!=targetColor){
                        if(left[color]<=j&&right[color]>=j&&top[color]<=i&&down[color]>=i){
                            if(next[color].find(targetColor)==next[color].end()){
                                next[color].insert(targetColor);
                                inDegree[targetColor]++;
                            }
                        }
                    }
                }
            }
        }

        // BFS
        queue<int> q;

        for(auto item: inDegree){
            int i=item.first, ide=item.second;
            if(ide==0){
                q.push(i);
            }
        }
        int count=0;

        while(!q.empty()){
            int cur = q.front();
            q.pop();
            count++;

            for(int next: next[cur]){
                inDegree[next]--;
                if(inDegree[next]==0){
                    q.push(next);
                }
            }
        }
        return count == countAll.size();
    }
};
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