<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 bfs(vector<vector<int>>& forest, int sr, int sc, int tr, int tc)
    {
        if(sr==tr && sc==tc)
            return 0;
        int m = forest.size();
        int n = forest[0].size();
        vector<vector<int>> vis(m,vector<int>(n,0));
        queue<pair<int,int>> q;
        q.push({sr,sc});
        vis[sr][sc] = 1;
        int steps = 0;
        int dr[] = {-1,0,0,1};
        int dc[] = {0,-1,1,0};
        while(!q.empty())
        {
            ++steps;
            int sz = q.size();
            for(int i = 0; i<sz; i++)
            {
                int r = q.front().first;
                int c = q.front().second;
                q.pop();
                for(int i = 0; i<4; i++)
                {
                    int nr = r+dr[i];
                    int nc = c+dc[i];
                    if(nr>=0 && nr<m && nc>=0 && nc<n && !vis[nr][nc] && forest[nr][nc]!=0)
                    {
                        if(nr==tr && nc==tc)
                            return steps;
                        vis[nr][nc] = 1;
                        q.push({nr,nc});    
                    }
                }
            }
        } 
        return -1;   
    }
    int cutOffTree(vector<vector<int>>& forest) {
        //int res = 0;
        int m = forest.size();
        int n = forest[0].size();
        vector<vector<int>> vis(m,vector<int>(n,0));
        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,greater<pair<int,pair<int,int>>>> pq;
        for(int i = 0; i<m; i++)
        {
            for(int j = 0; j<n; j++)
            {
                if(forest[i][j]>1)
                    pq.push({forest[i][j],{i,j}});
            }
        }
        int res = 0, sr = 0, sc = 0;
        while(!pq.empty())
        {
            auto it = pq.top();
            pq.pop();
            int tr = it.second.first;
            int tc = it.second.second;
            int steps = bfs(forest,sr,sc,tr,tc);
            if(steps==-1)
                return -1;
            res+=steps;
            sr = tr;
            sc = tc;    
        }
        return res;
    }
};

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