<p>You are asked to cut off all the trees in a forest for a golf event. The forest is represented as an <code>m x n</code> matrix. In this matrix:</p>

<ul>
	<li><code>0</code> means the cell cannot be walked through.</li>
	<li><code>1</code> represents an empty cell that can be walked through.</li>
	<li>A number greater than <code>1</code> represents a tree in a cell that can be walked through, and this number is the tree&#39;s height.</li>
</ul>

<p>In one step, you can walk in any of the four directions: north, east, south, and west. If you are standing in a cell with a tree, you can choose whether to cut it off.</p>

<p>You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes <code>1</code> (an empty cell).</p>

<p>Starting from the point <code>(0, 0)</code>, return <em>the minimum steps you need to walk to cut off all the trees</em>. If you cannot cut off all the trees, return <code>-1</code>.</p>

<p><strong>Note:</strong> The input is generated such that no two trees have the same height, and there is at least one tree needs to be cut off.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,4],[7,6,5]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Following the path above allows you to cut off the trees from shortest to tallest in 6 steps.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees2.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,0],[7,6,5]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The trees in the bottom row cannot be accessed as the middle row is blocked.
</pre>

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

<pre>
<strong>Input:</strong> forest = [[2,3,4],[0,0,5],[8,7,6]]
<strong>Output:</strong> 6
<b>Explanation:</b> You can follow the same path as Example 1 to cut off all the trees.
Note that you can cut off the first tree at (0, 0) before making any steps.
</pre>

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

<ul>
	<li><code>m == forest.length</code></li>
	<li><code>n == forest[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>0 &lt;= forest[i][j] &lt;= 10<sup>9</sup></code></li>
	<li>Heights of all trees are <strong>distinct</strong>.</li>
</ul>

class Solution {
public:
    int cutOffTree(vector<vector<int>>& forest) {
      int n = forest.size();
      int m = forest[0].size(); 
      vector <array <int, 3>> h;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
          if (forest[i][j] > 1) {
            h.push_back({forest[i][j], i, j});
          }
        }
      }
      sort(h.begin(), h.end());
      using a2 = array <int, 2>;
      auto valid = [&] (int x, int y) -> bool {
        return (x >= 0 && x < n && y >= 0 && y < m && forest[x][y] >= 1);
      };

      auto BFS = [&] (a2 src, a2 des) -> int {
        queue <a2> q;
        q.push(src);
        vector <vector <int>> dis (n, vector <int> (m, -1));
        dis[src[0]][src[1]] = 0;
        while(!q.empty()) {
          int sz = q.size();
          queue <a2> nq;
          while(sz--) {
            auto [x, y] = q.front();
            q.pop();
            for (int dx = -1; dx <= 1; dx++) {
              for (int dy = -1; dy <= 1; dy++) if (abs(dx) + abs(dy) == 1) {
                int nx = x + dx;
                int ny = y + dy;
                if (valid(nx, ny) && dis[nx][ny] == -1) {
                  dis[nx][ny] = dis[x][y] + 1;
                  nq.push({nx, ny});
                }
              }
            }
          }
          swap(q, nq);
        }
        auto [dx, dy] = des;
        return dis[dx][dy];
      };

      int res = 0;
      a2 cur = {0, 0};
      for (auto [ht, x, y] : h) {
        int ad = BFS(cur, {x, y});
        cout << ad << '\n';
        if (ad == -1) return -1;
        res += ad;
        cur = {x, y};
      }
      return res;
    }
};
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