<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 cur_x, cur_y;
    
    vector<int> trees_height;
    
    map<int, pair<int, int>> trees_coord;
    
    int steps = 0;
    
    const vector<vector<int>> dirs = {{-1,0}, {1,0}, {0,-1}, {0,1}};
    
    void init(vector<vector<int>>& forest) {
        for ( int i = 0; i < forest.size(); i++ ) {
            for ( int j = 0; j < forest[i].size(); j++ ) {
                if ( forest[i][j] > 1 ) {
                    trees_height.push_back(forest[i][j]);
                    trees_coord[forest[i][j]] = make_pair(i, j);
                }
            }
        }
        sort(trees_height.begin(), trees_height.end());
        cur_x = 0;
        cur_y = 0;
        steps = 0;
    }
    
    int run_bfs_to_tree(int tree_height, vector<vector<int>>& forest) {
        queue<pair<int, int>> bfs;
        bfs.push(make_pair(cur_x, cur_y));
        int end_x = trees_coord[tree_height].first;
        int end_y = trees_coord[tree_height].second;
        int cur_dist = 0;
        int end_coord_reached = false;
        vector<vector<bool>> visited = vector<vector<bool>>(forest.size(), vector<bool>(forest[0].size(), false));
        
        while ( !bfs.empty() ) {
            int n = bfs.size();
            
            while ( n > 0 ) {
                n--;
                int bfs_x_cord = bfs.front().first;
                int bfs_y_cord = bfs.front().second;
                bfs.pop();
                if ( visited[bfs_x_cord][bfs_y_cord] ) {
                    continue;
                }
                if ( bfs_x_cord == end_x && bfs_y_cord == end_y ) {
                    end_coord_reached = true;
                    return cur_dist;
                }
                visited[bfs_x_cord][bfs_y_cord] = true;
                for ( int i = 0; i < dirs.size(); i++ ) {
                    int next_x = bfs_x_cord + dirs[i][0];
                    int next_y = bfs_y_cord + dirs[i][1];
                    if ( next_x >= 0 && next_y >= 0 && next_x < forest.size() && next_y < forest[0].size() && forest[next_x][next_y] != 0 && !visited[next_x][next_y]) {
                        bfs.push(make_pair(next_x, next_y));
                    }
                }
            }
            cur_dist++;
        }
        return -1;
    }
    
    void run_bfs(vector<vector<int>>& forest) {
        for ( int i = 0; i < trees_height.size(); i++ ) {
            int steps_to_reach_here = run_bfs_to_tree(trees_height[i], forest);
            if ( steps_to_reach_here != -1 ) {
                steps += steps_to_reach_here;
                cur_x = trees_coord[trees_height[i]].first;
                cur_y = trees_coord[trees_height[i]].second;
                forest[cur_x][cur_y] = 1;
            }
            else {
                steps = -1;
                return;
            }
        }
    }
public:
    int cutOffTree(vector<vector<int>>& forest) {
        init(forest);
        run_bfs(forest);
        return steps;
    }
};
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