<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 Cell {
    int value;
    int row;
    int column;

    bool operator<(const Cell& other) const {
        return value > other.value; // Priority queue needs max-heap, so we reverse the comparison
    }
};

class Solution {
public:
    int cutOffTree(std::vector<std::vector<int>>& forest) {
        if (forest.empty() || forest[0].empty()) return -1;

        int m = forest.size();
        int n = forest[0].size();

        std::priority_queue<Cell> trees;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (forest[i][j] > 1) {
                    trees.push({forest[i][j], i, j});
                }
            }
        }

        int totalSteps = 0;
        int startX = 0, startY = 0;

        std::vector<std::vector<bool>> visited(m, std::vector<bool>(n, false));
        std::queue<std::pair<int, int>> q;

        while (!trees.empty()) {
            Cell tree = trees.top();
            trees.pop();
            int steps = bfs(forest, startX, startY, tree.row, tree.column, visited, q);
            if (steps == -1) return -1;
            totalSteps += steps;
            startX = tree.row;
            startY = tree.column;
        }

        return totalSteps;
    }

private:
    int bfs(const std::vector<std::vector<int>>& forest, int startX, int startY, int targetX, int targetY, std::vector<std::vector<bool>>& visited, std::queue<std::pair<int, int>>& q) {
        if (startX == targetX && startY == targetY) return 0;

        int m = forest.size();
        int n = forest[0].size();

        // Reset visited array and queue
        std::fill(visited.begin(), visited.end(), std::vector<bool>(n, false));
        std::queue<std::pair<int, int>> empty;
        std::swap(q, empty);

        q.push({startX, startY});
        visited[startX][startY] = true;

        int steps = 0;
        std::vector<int> directions = {-1, 0, 1, 0, -1};

        while (!q.empty()) {
            int size = q.size();
            ++steps;
            for (int i = 0; i < size; ++i) {
                auto [x, y] = q.front();
                q.pop();
                for (int d = 0; d < 4; ++d) {
                    int nx = x + directions[d];
                    int ny = y + directions[d + 1];
                    if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny] && forest[nx][ny] != 0) {
                        if (nx == targetX && ny == targetY) return steps;
                        q.push({nx, ny});
                        visited[nx][ny] = true;
                    }
                }
            }
        }

        return -1;
    }
};
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