<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 cutOffTree(vector<vector<int>>& forest) {
        std::vector<std::pair<int,int>> vec;
        for(int i = 0;i<forest.size();i++){
            for(int j = 0;j<forest[0].size();j++){
                if(forest[i][j]>1)vec.emplace_back(i,j);
            }
        }
        std::ranges::sort(vec,[&](auto a,auto b){
            return forest[a.first][a.second] < forest[b.first][b.second];
        });
        int acc = bfs(forest,vec);
        return acc<0 ? -1 : acc;
    }
private:
    int bfs(const vector<vector<int>>& forest,const std::vector<std::pair<int,int>>& vec,int i = 0,int j = 0,int curr = 0){
        if(curr>=vec.size())return 0;
        vector<vector<bool>> seen(forest.size(),std::vector<bool>(forest[0].size(),false));
        std::queue<std::pair<int,int>> que;//{{}}
        que.emplace(i,j);
        int k = 0;
        while(!que.empty()){
            int size = que.size();
            for(int index = 0;index<size;index++){
                auto [ni,nj] = que.front();
                que.pop();
                if(forest[ni][nj] == 0 || seen[ni][nj])continue;
                seen[ni][nj]=true;
                if(ni == vec[curr].first && nj == vec[curr].second){
                    return k+bfs(forest,vec,ni,nj,curr+1);
                }
                if(ni != 0)que.emplace(ni-1,nj);
                if(nj != 0)que.emplace(ni,nj-1);
                if(ni < forest.size()-1)que.emplace(ni+1,nj);
                if(nj < forest[0].size()-1)que.emplace(ni,nj+1);
            }
            ++k;
        }
        return INT_MIN;
    }
};
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