<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:
    map<int, int> hash;
    vector<int> DIR = {1, 0, -1, 0, 1};
    int n, m;
    int cutOffTree(vector<vector<int>>& f) {
        n = f.size(), m = f[0].size();
        int cur = 0, result = 0;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                if(f[i][j] > 1)
                    hash[f[i][j]] = i * m + j;
        
        for(auto& a: hash){
            int steps = helper(f, cur, a.second);
            if(steps == -1e9) return -1;
            result += steps;
            cur = a.second;
        }
        return result;
    }
    
    int helper(vector<vector<int>>& f, int cur, int target){
        queue<int> q;
        int steps = 0;
        q.push({cur});
        vector<vector<bool>> vis(n, vector<bool>(m));
        while(!q.empty()){
            int sz = q.size();
            while(sz--){
                int pos = q.front(), r = pos / m, c = pos % m;
                q.pop();
                if(pos == target)
                    return steps;
                for(int k = 0; k < 4; k++){
                    int nr = r + DIR[k], nc = c + DIR[k + 1];
                    if(nr < 0 || nr == n || nc < 0 || nc == m || vis[nr][nc] || !f[nr][nc])
                        continue;
                    q.push({nr * m + nc});
                    vis[nr][nc] = true;
                }
            }
            steps++;
        }
        return -1e9;
    }
};
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