<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 solve(vector<vector<int>>&arr,int i,int j,int s,int e,int n,int m)
    {
        queue<pair<int,pair<int,int>>>q;
        q.push({i,{j,0}});
        vector<vector<int>>vis(n,vector<int>(m,0));
        vis[i][j]=1;
        while(!q.empty()){
            auto p=q.front();
            q.pop();
            int x=p.first,y=p.second.first,dis=p.second.second;
            if(x==s && y==e)
             return dis;

            if(x+1<n && arr[x+1][y]!=0 && vis[x+1][y]==0)
            {
                vis[x+1][y]=1;
                if(x+1==s && y==e)
                 return dis+1;
                q.push({x+1,{y,dis+1}});
            } 
            if(x-1>=0 && arr[x-1][y]!=0 && vis[x-1][y]==0)
            {
                vis[x-1][y]=1;
                if(x-1==s && y==e)
                 return dis+1;
                q.push({x-1,{y,dis+1}});
            } 
            if(y+1<m && arr[x][y+1]!=0 && vis[x][y+1]==0)
            {
                vis[x][y+1]=1;
                if(x==s && y+1==e)
                 return dis+1;
                q.push({x,{y+1,dis+1}});
            } 
            if(y-1>=0 && arr[x][y-1]!=0 && vis[x][y-1]==0)
            {
                vis[x][y-1]=1;
                if(x==s && y-1==e)
                 return dis+1;
                q.push({x,{y-1,dis+1}});
            } 
        }
        return -1;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        vector<vector<int>>arr;
        int n=forest.size();
        int m=forest[0].size();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(forest[i][j]>1)
                    arr.push_back({i,j});
            }
        }
        sort(arr.begin(),arr.end(),[&](vector<int>&a,vector<int>&b){
            return forest[a[0]][a[1]]<forest[b[0]][b[1]];
        });

        // for(auto x:arr)
        //  cout<<x[0]<<":"<<x[1]<<" ";
        int i=0;
        int j=0;
        int ans=0;
        for(auto x:arr)
        {
            int s=x[0];
            int e=x[1];
            int dis=solve(forest,i,j,s,e,n,m);
            if(dis==-1)
             return -1;

            ans+=dis;
            i=s;
            j=e; 
        }

        return ans; 
    }
};
303685