<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:
    bool bfs(int find,pair<int,int> &start,vector<vector<int>> &f,int &total)
    {
        int m=f.size(),n=f[0].size();
        int steps=0;
        queue<pair<int,int>> q;
        vector<vector<bool>> vis(m,vector<bool>(n,false));
        q.push(start);
        vis[start.first][start.second]=true;
        int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
        while(!q.empty())
        {
            int sz=q.size();
            steps++;
            for(int i=0;i<sz;i++)
            {
                pair<int,int> pos=q.front();q.pop();
                for(int i=0;i<4;i++)
                {
                    int nx=pos.first+dx[i];
                    int ny=pos.second+dy[i];
                    if(nx>=0&&ny>=0&&nx<m&&ny<n&&f[nx][ny]!=0&&!vis[nx][ny])
                    {
                        if(f[nx][ny]==find)
                        {
                            total+=steps;
                            start={nx,ny};
                            f[nx][ny]=1;
                            return true;
                        }
                        else if(f[nx][ny]>0) q.push({nx,ny});
                        vis[nx][ny]=1;
                    }
                }
            }
        }
        return false;
    }
    int cutOffTree(vector<vector<int>>& f) {
        int m=f.size(),n=f[0].size();
        priority_queue<int,vector<int>,greater<int>> order;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                if(f[i][j]>1) order.push(f[i][j]);
        pair<int,int> start={0,0};
        int total=0;
        while(order.size()>0)
        {
            int next=order.top();
            if(f[start.first][start.second]==next)
            {
                order.pop();
                f[start.first][start.second]=1;
                continue;
            }
            if(!bfs(next,start,f,total)) return -1;
            order.pop();
        }
        return total;
    }
};
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