<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>

struct TreePos {
    int height, r, c;
    bool operator==(const TreePos & other) const {
        return height == other.height;
    }
    bool operator<(const TreePos & other) const {
        return height < other.height;
    }
};

class Solution {
    int m, n;
    const int dirs[4][2] = {
        {-1, 0},
        {1, 0},
        {0, 1},
        {0, -1}
    };

    int next_step(vector<vector<int>>& forest, TreePos start, TreePos end) {
        if (start.r == end.r && start.c == end.c)
            return 0;

        queue<TreePos> q;
        q.push(start);

        vector<vector<bool>> visited(m, vector<bool>(n, false));
        visited[start.r][start.c] = true;

        int step = 0;

        while (!q.empty()) {
            ++step;
            int q_size = q.size();
            while (q_size--) {
                auto top = q.front();
                q.pop();
                
                for (auto & dir: dirs) {
                    int r = top.r + dir[0], c = top.c + dir[1];
                    if (r < 0 || r == m || c < 0 || c == n || visited[r][c] || !forest[r][c])
                        continue;

                    TreePos next = { forest[r][c], r, c };
                    if (next == end)
                        return step;
                    
                    visited[r][c] = true;
                    q.push(next);
                }
            }
        }

        return -1;
    }
public:
    int cutOffTree(vector<vector<int>>& forest) {

        m = forest.size(), n = forest[0].size(); 
        vector<TreePos> trees;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (forest[i][j] > 1)
                    trees.push_back({ forest[i][j], i, j });
            }
        }

        sort(trees.begin(), trees.end());

        int result = 0, num_trees = trees.size();
        TreePos start = { forest[0][0], 0, 0 };

        for (int i = 0; i < num_trees; ++i) {
            int step = next_step(forest, start, trees[i]);
            if (step == -1)
                return -1;
            result += step;
            start = trees[i];
        }

        return result;
    }
};
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