<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 vis[54][54];
    void fill_v()
    {
        for(int i =0;i<54;i++)
        {
            for(int j=0;j<54;j++)
            {
                vis[i][j] = 0;
            }
        }
    }
    
    
    int bfs(int si, int sj, int ei, int ej,vector<vector<int>> &f)
    {
        queue<pair<int,pair<int,int>>> q;
        q.push({0,{si,sj}});
        int ans = -1;
        
        int n = f.size();
        int m = f[0].size();
        
        while(!q.empty())
        {
            int level = q.front().first;
            int i = q.front().second.first;
            int j = q.front().second.second;
            
            q.pop();
            if(vis[i][j])
                continue;
            
            vis[i][j] = 1;
            
            
            
            if(i == ei && j == ej)
            {
                return level;
            }
            
            if(i+1 < n && j<m && vis[i+1][j] == 0 && f[i+1][j])
            {
                q.push({level+1, {i+1, j}});
            }
            
            if(i < n && j+1<m && vis[i][j+1] == 0 && f[i][j+1])
            {
                q.push({level+1, {i, j+1}});
            }
            
            if(i-1>=0 && i-1 < n && j<m && vis[i-1][j] == 0 && f[i-1][j])
            {
                q.push({level+1, {i-1, j}});
            }
            
            if(j-1>=0 && i< n && j-1<m && vis[i][j-1] == 0 && f[i][j-1])
            {
                q.push({level+1, {i, j-1}});
            }
        }
        
        return ans;
            
    }
    
    int cutOffTree(vector<vector<int>>& f) {
        
        
        int n = f.size();
        int m = f[0].size();
        vector<pair<int,pair<int,int>>> v; 
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(f[i][j]>1)
                    v.push_back({f[i][j],{i,j}});
            }
            
        }
        
        v.push_back({0,{0,0}});
        int l = v.size();
        
        sort(v.begin(),v.end());
        int ans = 0;
        for(int i=0;i<l-1;i++)
        {
            fill_v();
            int si,sj,ei,ej;
            si = v[i].second.first;
            sj = v[i].second.second;
            
            ei = v[i+1].second.first;
            ej = v[i+1].second.second;
            
            int res = bfs( si, sj,  ei, ej,f);
            if(res == -1)
                return res;
            ans+=res;
        }
        
        return ans;
    }
};
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