<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 dx[4]={1,0,-1,0};
    int dy[4]={0,1,0,-1};
    int l;
    void bfs(int x,int y,int &r,int &c,vector<vector<bool>>&vis,vector<vector<int>>&level)
    {
        queue<pair<int,int>>q;
        vis[x][y]=1;
        q.push({x,y});
        level[x][y]=0;
        while(q.size())
        {
           auto a=q.front();
           q.pop();
           for(int i=0;i<4;i++)
           {
            int nx=a.first+dx[i];
            int ny=a.second+dy[i];
            if(nx<0||ny<0||nx>=r||ny>=c||vis[nx][ny])
            {
                continue;
            }
            else
            {
                vis[nx][ny]=1;
                q.push({nx,ny});
                level[nx][ny]=level[a.first][a.second]+1;

            }
           }
        }
    }
    bool check(int &r,int &c,vector<vector<bool>>&vis,vector<vector<int>>&forest,vector<vector<int>>&level,int p)
    {
                for(int i=0;i<r;i++)
                {
                    for(int j=0;j<c;j++)
                    {
                        if(vis[i][j]==0)
                        {
                            
                            return 0;
                        }
                        else if(forest[i][j]==p)
                        {
                           l=level[i][j];
                        }
                    }
                }
             return 1;
    }
    int cutOffTree(vector<vector<int>>& forest) 
    {
        int r=forest.size();
        int c=forest[0].size();
        int INF=1e9+7;
        vector<int>v;
        vector<vector<bool>>vis(r,vector<bool>(c,0)),vis1;
        vector<vector<int>>level(r,vector<int>(c,INF)),level1;
     
        for(int i=0;i<r;i++)
        {
            for(int j=0;j<c;j++)
            {
                if(forest[i][j]==0)
                 {
                   
                    vis[i][j]=1;
                 }
                 else
                 {
                    if(forest[i][j]>1)
                    {
                    v.push_back(forest[i][j]);
                    }
                  
                 }
            }
        }
        level1=level;
        vis1=vis;
        sort(v.begin(),v.end());
        int ans=0;
        if(v.size()>1)
        {
    for(int t=0;t<=v.size()-2;t++)
     {
                        
        for(int i=0;i<r;i++)
        {
            for(int j=0;j<c;j++)
            {
                if(vis[i][j]==0)
                {
                 
                    if(forest[i][j]==v[t])
                    {
                    
                        bfs(i,j,r,c,vis,level); 
                        if(!check(r,c,vis,forest ,level,v[t+1]))
                        {
                           return -1;
                        }
                        else
                        {
                            ans=ans+(l);
                            
                        }
                        
                        level=level1;
                        vis=vis1;
                        
                       
                    }
                    

                }
            }
        }
     }
        }
     
        level=level1;
        vis=vis1;
        if(forest[0][0]!=0)
        {
        bfs(0,0,r,c,vis,level);
        }
        else
        {
            return -1;
        }
        for(int i=0;i<r;i++)
        {
            for(int j=0;j<c;j++)
            {
                if(forest[i][j]==v[0])
                {
                    ans=ans+level[i][j];
                    break;
                }
            }
        }
     return ans;
            
        
    }
};
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