<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 i,j;
    int dfs(int d, vector<vector<int>>& forest){
        queue<pair<int, int>>q;
        int n,m;
        n=forest.size();
        m=forest[0].size();
        int dist[51][51];
        for(int k=0;k<n;k++)
        for(int z=0;z<m;z++)
            dist[k][z]=-1;
        q.push({i,j});
        dist[i][j]=0;
        while(!q.empty()){
            pair<int,int>p = q.front();
            q.pop();
            if(forest[p.first][p.second]==d){
                i=p.first;
                j=p.second;
                return dist[i][j];
            }
            if(p.first-1>=0&&forest[p.first-1][p.second]!=0&&dist[p.first-1][p.second]==-1){
                dist[p.first-1][p.second] = dist[p.first][p.second] +1;
                q.push({p.first-1,p.second});
            }
            if(p.first+1<n&&forest[p.first+1][p.second]!=0&&dist[p.first+1][p.second]==-1){
                dist[p.first+1][p.second] = dist[p.first][p.second] +1;
                q.push({p.first+1,p.second});
            }
            if(p.second-1>=0&&forest[p.first][p.second-1]!=0&&dist[p.first][p.second-1]==-1){
                dist[p.first][p.second-1] = dist[p.first][p.second] +1;
                q.push({p.first,p.second-1});
            }
            if(p.second+1<m&&forest[p.first][p.second+1]!=0&&dist[p.first][p.second+1]==-1){
                dist[p.first][p.second+1] = dist[p.first][p.second] +1;
                q.push({p.first,p.second+1});
            }
        }
        return -1;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        if(forest[0][0]==0){
            return -1;
        }
        vector<int>v;
        i=j=0;
        for(int k=0;k<forest.size();k++){
            for(int n=0;n<forest[k].size();n++){
                if(forest[k][n]>1)
                    v.push_back(forest[k][n]);
            }
        }
        sort(v.begin(),v.end());
        int ans = 0;
        for(int k=0;k<v.size();k++){
            int x = dfs(v[k], forest);
            if(x==-1)
                return -1;
            ans+=x;
        }
        return ans;
    }
};
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