<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 dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
    int solve(int s,int e,int x,int y,vector<vector<int>>& forest){
    if(s==x && e==y) return 0;
    
    int n=forest.size();
    int m=forest[0].size();
    queue<pair<int,int>> q;
    int vis[n][m];
    memset(vis,0,sizeof(vis));
    
    vis[s][e]=1;
    q.push({s,e});
    int count=0;
    while(!q.empty()){
        count++;
        int nn=q.size();
        while(nn--){
            auto it=q.front();
            q.pop();
            
            for(int i=0;i<4;i++){
                int n_x=it.first+dir[i][0];
                int n_y=it.second+dir[i][1];
                if(n_x<0 || n_y<0 || n_x>=n || n_y>=m || vis[n_x][n_y]==1 || forest[n_x][n_y]==0){
                    continue;
                }
                q.push({n_x,n_y});
                vis[n_x][n_y]=1;
                if(n_x==x && n_y==y) return count;
               
            }
        }
    }
    return -1;
}

int cutOffTree(vector<vector<int>>& forest) {
    vector<vector<int>>ans;
    for(int i=0;i<forest.size();i++){
        for(int j=0;j<forest[i].size();j++){
            if(forest[i][j]>1){
                ans.push_back({forest[i][j],i,j});
            }
        }
    }
    sort(ans.begin(),ans.end());
    int s=0,e=0;
    int res=0;
    for(auto &it:ans){
        int d=solve(s,e,it[1],it[2],forest);
        if(d==-1){
            return -1;
        }
        res+=d;
        s=it[1];
        e=it[2];
    }
    return res;
    
}
};
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