<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 n,m;
    vector<vector<int>> map;
    vector<int> dist;
   // vector<int> visited;
    vector<int> toCoord(int x){
        vector<int> ans(2);
        ans[0] = x / n;
        ans[1] = x % n;
        return ans;
    }
    int toIdx(int i, int j){
        return i*n+ j;
    }
    void add2q(int k, queue<int> & myQ, vector<int> &dist_k){
        const vector<vector<int>> step = {{0,1},{1,0},{0,-1},{-1,0}};
        vector<int> x = {k/n, k%n};

        for (auto & d : step){
            int i = x[0]+d[0];
            int j = x[1]+d[1];
            int idx = i*n+j;
           // cout << i << " " << j << " " << idx <<endl;
            if (i>=0 && i<m && j>=0 && j<n )
              if ( map[i][j]>0 ){
                if (dist_k[idx] == -1 ){
                    dist_k[idx] = dist_k[k] + 1;
                    myQ.push(idx);
                }
            }
        }

    }
    int bfs(int a, int b, int c){
        
       //      cout << a << " " << b << " " << c <<endl;
            queue<int> myQ;
        //    dist.resize(n*m);
            dist.assign(n*m, -1);
            dist[a] = 0;
           const vector<vector<int>> step = {{0,1},{1,0},{0,-1},{-1,0}};
            myQ.push(a);
            int i, j, idx;
            while (!myQ.empty()){
                int t = myQ.front();
                myQ.pop();
               // add2q(t,myQ, dist);
                
                int k = t; 
        vector<int> x = {k/n, k%n};

        for (auto & d : step){
             i = x[0]+d[0];
            j = x[1]+d[1];
            idx = i*n+j;
           // cout << i << " " << j << " " << idx <<endl;
            if (i>=0 && i<m && j>=0 && j<n )
              if ( map[i][j]>0 ){
                if (dist[idx] == -1 ){
                    dist[idx] = dist[k] + 1;
                    myQ.push(idx);
                }
            }
        }

             //   cout << myQ.size() << endl;
                if ( b>-1 && dist[b] > -1 && (c== -1 || c>-1&&dist[c] > -1) )
                    return dist[b];
              //  if (found)
               //     return dist[b];
                
            }
            
            if (b > -1)
                return dist[b];
            else{
                for (int k = 0; k<n*m; k++)
                    if (dist[k] == -1 && map[k / n][k%n] >0)
                        return -1;
                return 0;
            }

        
    }
    int cutOffTree(vector<vector<int>>& forest) {
        map = forest;
        m = map.size();
        n = map[0].size();
         dist.resize(n*m);
        priority_queue<pair<int,int>> list;
        for (int i = 0; i<forest.size(); i++){
            for (int j = 0; j<forest[0].size(); j++){
                if (forest[i][j]>1){
                    list.push( make_pair(-forest[i][j],toIdx(i,j)) );
                }
            }
        }
        //auto prev =  make_pair(0,0);
        int a = 0, b,c;
        int ans = 0, d = 0;
        bool redo_bfs = true;
   //     cout << "***"<< endl;
        if (bfs(0,-1, -1) == -1) return -1;
     //   cout << "***"<< endl;
        while (!list.empty()){
            //auto next = list.top();
            b = list.top().second;
            list.pop();
            if (!list.empty()){
                c = list.top().second;
                list.pop();
            }
            else c = -1;

       //     if (next.first == -1) continue;
            //cout << prev.second << " " << next.second <<endl;
           
            d =  bfs(b,a,c);
            if (d == -1) return -1;
            ans += d;

            if (c>-1) 
            {
                d = dist[c];
                if (d == -1) return -1;
                ans += d;
            }
         //  cout << ans << " "<<list.size() << " "<<d <<endl;
          //  ans++;
          //  list.pop();
          //  prev = next;
          //  redo_bfs = !redo_bfs;
          a = c;
          
        }
        return ans;
    }
};
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