<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:
    static bool comp(pair<int,int> a, pair<int,int> b){
        return a.first < b.first;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        
        int n = forest.size();
        int m = forest[0].size();
        
        vector<vector<int>> visited(n,vector<int>(m,0));
        vector<vector<int>> ref(n,vector<int>(m,0));

        vector<pair<int,int> > order;
        for(int i = 0; i < forest.size();i++){
            for(int j = 0; j < forest[i].size();j++){
                if(forest[i][j] >1){
                    order.push_back(make_pair(forest[i][j],i*m+j));
                }
            }
        }
        sort(order.begin(),order.end(),comp);
        
        
        
        
        queue<int> dfs;
        dfs.push(0);
        int steps = 0;
        int amtPerLevel = 1;
        int treeNum = 0;
        int curSteps = 0;
        visited[0][0] = 1;
        while(treeNum < order.size()){
            
            if(dfs.size() == 0){
                return -1;
            }
            
            int top = dfs.front();
            dfs.pop();
            amtPerLevel--;
            
            int i = top/m;
            int j = top%m;
            // cout <<i << " " << j << endl;

            // [4,2,3],
            // [0,0,1],
            // [7,6,5]
            
            if(order[treeNum].first == forest[i][j]){
                
                dfs= {};
                if(treeNum < order.size()){
                    dfs.push(order[treeNum].second);
                }
                treeNum++;
                visited = ref;
                // cout << "curSteps" <<  curSteps << endl;

                amtPerLevel = 1;
                steps+=curSteps;
                curSteps = 0;
                continue;
            }

            bool isZero = amtPerLevel == 0;
            if(i-1>=0 && !visited[i-1][j] && forest[i-1][j] != 0){
                dfs.push((i-1)*m+j);
                visited[i-1][j] = 1;
            }
            if(i+1 < n&& !visited[i+1][j] && forest[i+1][j] != 0){
                dfs.push((i+1)*m+j);
                visited[i+1][j] = 1;
            }
            if(j-1>=0&& !visited[i][j-1] && forest[i][j-1] != 0){
                dfs.push(i*m+j-1);
                visited[i][j-1] = 1;
            }
            if(j+1 < m&& !visited[i][j+1] && forest[i][j+1] != 0){
                dfs.push(i*m+j+1);
                visited[i][j+1] = 1;
            }
            if(isZero){
                amtPerLevel = dfs.size();
                curSteps++;
            }
            
            
        }
        return steps;
    }
};
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