<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:
    def cutOffTree (self, forest: List[List[int]]) -> int:
        ROWS, COLS = len(forest), len(forest[0])
        minHeap = []
        heapq.heapify(minHeap)
        reuse_visited = [[0 for _ in range(COLS)] for _ in range(ROWS)]
        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        for i in range(ROWS):
            for j in range(COLS):
                if forest[i][j] == 0:
                    reuse_visited[i][j] = 1
                elif forest[i][j] > 1:
                    heapq.heappush(minHeap, (forest[i][j], i, j))

        def BFS(src, target):
            queue = [(src[0], src[1], 0)] 
            visited = [row[:] for row in reuse_visited] 
            visited[src[0]][src[1]] = 1
            while queue:
                tree_i, tree_j, step = queue.pop(0)
                for delrow, delcol in directions:
                    new_i, new_j = tree_i + delrow, tree_j + delcol
                    if 0 <= new_i < ROWS and 0 <= new_j < COLS and not visited[new_i][new_j]:
                        if (new_i, new_j) == target:
                            return step + 1
                        queue.append((new_i, new_j, step+1))
                        visited[new_i][new_j] = 1
            return -1
        
        src = (0, 0)
        res = 0
        while minHeap:
            tree_height, tree_i, tree_j = heapq.heappop(minHeap)
            if (tree_i, tree_j) == (src[0], src[1]):
                tree_height, tree_i, tree_j = heapq.heappop(minHeap)
            temp = BFS(src, (tree_i, tree_j))
            if temp == -1:
                return -1
            res += temp
            src = (tree_i, tree_j)

        return res
16800