<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 {
public:
    
    // directions: up, down, left, right
    const vector<pair<int,int>> directions = {{-1,0}, {1,0}, {0,-1}, {0,1}};
    
    //BFS function to go from (sr, sc) to (tr, tc)
    int bfs(const vector<vector<int>>& forest, int sr, int sc, int tr, int tc) {
       
        if (sr == tr && sc == tc) {
            return 0;
        }
        
        int m = forest.size();
        int n = forest[0].size();
        
        vector<vector<bool>> visited(m, vector<bool>(n,false));
        queue<pair<int,int>> q;
        q.emplace(sr, sc);
        visited[sr][sc] = true;
        int steps = 0;
        
        while (!q.empty()) {
            int q_size = q.size();
            steps++;
            for (int i=0; i<q_size; ++i) {
                auto[r,c] = q.front();
                q.pop();
                for (auto& dir : directions) {
                    int nr = r + dir.first;
                    int nc = c + dir.second;
                    if (nr >= 0 && nr < m && nc >= 0 && nc < n && !visited[nr][nc] && forest[nr][nc] != 0) {
                        if (nr == tr && nc == tc) return steps;
                        q.emplace(nr, nc);
                        visited[nr][nc] = true;
                        
                    }
                }
            }
        }
        return -1; // target not reachable
    }
    
    
    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size();
        if (m == 0) return -1;
        int n = forest[0].size();
        
        
        vector<tuple<int, int, int>> trees; //height, row, col
        for (int r = 0; r<m ; ++r) {
            for (int c = 0; c < n; ++c) {
                if (forest[r][c] > 1)
                    trees.emplace_back(forest[r][c], r, c);
                
            }
        }
        
        sort(trees.begin(), trees.end(), [&](const tuple<int, int, int>& a, const tuple<int, int, int>& b) -> bool { return get<0>(a) < get <0>(b);                                                                                              });
        
        int total_steps = 0;
        int sr = 0, sc = 0;
        
        for(auto& tree: trees) {
            int height, tr, tc;
            tie(height,tr,tc) = tree;
            int steps = bfs(forest, sr, sc, tr, tc);
            if (steps == -1) return -1;
            total_steps += steps;
            sr=tr;
            sc=tc;
        }
        return total_steps;
    }
};
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