<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 {
private:
    int hashCoord(int x, int y){
        return 50*x+y;
    }
public:
    int cutOffTree(const vector<vector<int>>& forest) {
        // must be from shortest to tallest
        // -1 if impossible
        // can backtrack if needed
        // all tree heights distinct! = a clear ordering
        // shortest path to all other nodes? -> bfs at each step
        vector<int> nn;
        for(int i = 0; i < forest.size(); ++i){
            for(int j : forest[i]) if(j>1) nn.push_back(j);
        }
        sort(nn.begin(), nn.end()); // small to large trees
        // now perform repeated bfs from small to large
        deque<pair<int,int>> bfsQueue;
        const vector<int> dirX = {0, 1, 0, -1};
        const vector<int> dirY = {1, 0, -1, 0};
        int ans = 0, goal = 0, dist = 0;
        bfsQueue.push_back({0,0});
        unordered_set<int> seen;
        seen.insert(hashCoord(0,0));
        deque<pair<int,int>> nQ;

        while(goal < nn.size() && !bfsQueue.empty()){
            ++dist;
            // empty current layer
            while(!bfsQueue.empty()){
                // perform bfs to reach the goal
                int x = bfsQueue.front().first;
                int y = bfsQueue.front().second;
                int xyhash = hashCoord(x,y);
                bfsQueue.pop_front();
                // cout << x << "," << y << " || ";
                // case: found the shortest
                if(forest[x][y] == nn[goal]){
                    ++goal;
                    ans += dist-1; // bfs is min dist;
                    nQ = {{x,y}};
                    seen = {xyhash};
                    dist = 0;
                    break;
                }
                // enqueue the next set
                else{
                    for(int i = 0; i < 4; ++i){
                        int nX = x + dirX[i];
                        int nY = y + dirY[i];
                        int hash = hashCoord(nX, nY);
                        if(
                           (nX > -1 && nX < forest.size())
                           && (nY > -1 && nY < forest[0].size())
                           && !seen.count(hash)
                           && forest[nX][nY] != 0
                          ){
                            nQ.push_back({nX,nY});
                            seen.insert(hash);
                        }
                    }
                }
            }
            bfsQueue.swap(nQ);
            nQ.clear();
        }
        if(goal < nn.size()) return -1;
        return ans;
    }
};
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