<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,vector<int> src, vector<int> des)
    {
        int n=forest.size();
        int m=forest[0].size();
        int sr = src[0];
        int sc = src[1];
        int dr = des[0];
        int dc = des[1];

        vector<vector<int>> vis(n,vector<int>(m,0));
        queue<pair<pair<int,int>,int>> q;
        if(forest[sr][sc]==0)
            return -1;
        q.push({{sr,sc},0});
        int dx[4] = {-1,1,0,0};
        int dy[4] = {0,0,-1,1};
        vis[sr][sc]=1;
        // cout<<dr<<" "<<dc;
        while(!q.empty())
        {
            auto c = q.front();
            q.pop();
            int i=c.first.first;
            int j=c.first.second;
            int step = c.second;
            // cout<<i<<" "<<j<<"bfs"<<endl;
            if(i==dr and j==dc)
                return step;
            for(int k=0;k<4;k++)
            {
                int x = i+dx[k];
                int y = j+dy[k];
                if(x>=0 and x<n and y>=0 and y<m and forest[x][y]>=1 and vis[x][y]==0)
                {
                    vis[x][y]=1;
                    q.push({{x,y},step+1});
                }
            }
        }
        return -1;
    }
    int cutOffTree(vector<vector<int>>& forest) {
        map<int, vector<int>> trees;

        // find location of all trees to cut
        for(int i = 0; i < forest.size(); i++){
            for(int j = 0; j<forest[0].size(); j++){
                if(forest[i][j] > 1) trees[forest[i][j]] = {i,j};
            }
        }
        int ans = 0;
        vector<int> curr{0,0};
        while(!trees.empty())
        {
            vector<int> small = trees.begin()->second;
            trees.erase(trees.begin());
            int distance = bfs(forest,curr,small);
            // cout<<distance<<" "<<curr[0]<<" "<<curr[1]<<" "<<small[0]<<" "<<small[1]<<endl;
            if(distance<0) return -1;
            ans+=distance;
            curr = small;
        }
        return ans;
    }
};
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