<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:
    typedef pair<int,int> P;
    vector<vector<int>>dist;
    vector<int>vec;
    int x,y;

    int f(int i, int j , vector<vector<int>>&grid){
        int n = grid.size();
        int m = grid[0].size();
        
        int steps=0;
        
        queue<pair<int,int>>q;
        q.push({i,j});
        int drow[4]={1,0,-1,0};
        int dcol[4]={0,1,0,-1};
        dist[i][j]=0;
        while(!q.empty()){
            int row = q.front().first;
            int col = q.front().second;
            int val = grid[row][col];
            q.pop();
            if(val==*vec.begin()){
                // cout<<val<<endl;
                cout<<dist[row][col]<<endl;
                // cnt+=dist[row][col];
                vec.erase(vec.begin());
                steps+=dist[row][col];
                while(!q.empty())q.pop();
                dist.assign(n,vector<int>(m,1e9));
                dist[row][col]=0;
                grid[row][col]=1;
                if(vec.size()==0)return steps;
                
            }
            
            for(int k=0;k<4;k++){
                int nr = row+drow[k];
                int nc = col+dcol[k];

                if(nr>=0 && nr<n && nc>=0 && nc<m && grid[nr][nc]>0){
                   if(dist[nr][nc]>dist[row][col]+1) {
                    dist[nr][nc]=1+dist[row][col];
                    q.push({nr,nc});
                   }
                }

            }
            
        }
        
        return steps;

    }
    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size();
        int m = forest[0].size();
        if(forest[0][0]==0)return -1;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int val = forest[i][j];
                if(val>1){
                    vec.push_back(val);
                }
            }
        }
        sort(vec.begin(),vec.end());
        dist.resize(n,vector<int>(m,1e9));
        int ans = f(0,0,forest);
        // cout<<vec.size()<<endl;
        if(vec.size()==0)return ans;
        return -1;
      
    }
};
90195