<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:
    int bfs(vector<vector<int>>& forest, int x1, int y1, int x2, int y2) {
        int row = forest.size();
        int col = forest[0].size();
        
        vector<vector<int>> visited(row, vector<int>(col, 0));
        
        int dx[] = {-1, 1, 0, 0};
        int dy[] = {0, 0, -1, 1};

        if(x1 == x2 && y1 == y2)
        return 0;
        
        int ans = 0;
        
        queue<pair<int, int>> q;
        
        q.push({x1, y1});
        visited[x1][y1] = 1;
        
        while (!q.empty()) {
            int size = q.size();
            
            while (size--) {
                int x = q.front().first;
                int y = q.front().second;
                q.pop();
                
                for (int i = 0; i < 4; i++) {
                    int nx = x + dx[i];
                    int ny = y + dy[i];
                    
                    if (nx == x2 && ny == y2)
                        return ans + 1;
                    
                    if (nx < 0 || nx >= row || ny < 0 || ny >= col || visited[nx][ny] || forest[nx][ny] == 0)
                        continue;
                    
                    visited[nx][ny] = 1;
                    q.push({nx, ny});
                }
            }
            ans++;
        }
        
        return -1;
    }
    
    int cutOffTree(vector<vector<int>>& forest) {
        if(forest.empty() || forest[0].empty())
            return -1;

        vector<pair<int, pair<int, int>>> tree;
        int row = forest.size();
        int col = forest[0].size();
        
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (forest[i][j] > 1)
                    tree.push_back({forest[i][j], {i, j}});
            }
        }
        
        sort(tree.begin(), tree.end());
        
        int ans = 0;
        
        int startX = 0, startY = 0;
        for (auto& t : tree) {
            int targetX = t.second.first;
            int targetY = t.second.second;
            
            int steps = bfs(forest, startX, startY, targetX, targetY);
            if (steps == -1)
                return -1;
            
            ans += steps;
            startX = targetX;
            startY = targetY;
        }
        
        return ans;
    }
};

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