<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>

using i32 = int;
class Solution {
public:
    bool isPrintable(vector<vector<int>>& targetGrid) {
      const i32 K = 60;
      array <i32, 4> _def = {K + 1, K + 1, 0, 0};
      vector <array <i32, 4>> A (K + 1, _def);
      auto Do = [&] (i32 &x, i32 y, bool mx = 1) -> void {
        if (mx == 1) x = max(x, y);
        else x = min(x, y);
      };
      auto Adj = [&] (i32 i, i32 j) -> void {
        i32 x = targetGrid[i][j];
        Do(A[x][0], i, 0);
        Do(A[x][1], j, 0);
        Do(A[x][2], i);
        Do(A[x][3], j);
      };
      
      i32 N = targetGrid.size();
      i32 M = targetGrid[0].size();
      vector <i32> vis (K + 1, 1);
      for (i32 i = 0; i < N; ++i) {
        for (i32 j = 0; j < M; ++j) {
          Adj(i, j);
          vis[targetGrid[i][j]] = 0;
        }
      }
      // for (i32 i = 0; i < N; ++i) {
      //   for (i32 j = 0; j < M; ++j) {
      //     i32 x = targetGrid[i][j];
      //     cout << A[x][0] << ' ' << A[x][1] << ' ' << A[x][2] << ' ' << A[x][3] << '\n';
      //   }
      // }
      
      i32 cnt = 0;
      auto chk = [&] (i32 x) -> bool {
        for (i32 i = A[x][0]; i <= A[x][2]; ++i) {
          for (i32 j = A[x][1]; j <= A[x][3]; ++j) {
            if (targetGrid[i][j] != x && targetGrid[i][j] != 0) return 0;
          }
        }
        return 1;
      };
      auto Relax = [&] () -> bool {
        for (i32 x = 1; x <= K; x++) if (!vis[x] && chk(x)) {
          vis[x] = 1;
          for (i32 i = A[x][0]; i <= A[x][2]; ++i) {
            for (i32 j = A[x][1]; j <= A[x][3]; ++j) {
              cnt += (targetGrid[i][j] != 0);
              targetGrid[i][j] = 0;
            }
          }
          return 1;
        }
        return 0;
      };

      while (Relax()) {}
      if (cnt == N * M) return true;
      return false;
    }
};
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