<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:
    map<int, int> hm;
    vector<vector<int>> dir = {{0,1}, {0,-1}, {-1, 0}, {1,0}};
    int n, m;

    int cutOffTree(vector<vector<int>>& forest) {
        n = forest.size();
        m = forest[0].size();
        int cur = 0;
        vector<vector<int>> trees;
        for(int i=0;i<n;i++) {
            for(int j=0;j<m;j++) {
                if(forest[i][j]>1) {
                    trees.push_back({forest[i][j], i, j});
                }
            }
        }
        sort(trees.begin(), trees.end());
        int sx = 0, sy=0;
        int ans =0;
        for(int i=0;i<trees.size();i++) {
            int cost = bfs(forest, sx, sy, trees[i][1], trees[i][2]);
            if(cost ==-1)
            return -1;
            ans+=cost;
            sx = trees[i][1];
            sy = trees[i][2];
        }
        return ans;
    }

    int bfs(vector<vector<int>> forest, int sx, int sy, int fx, int fy) {
        if(fx == sx && fy==sy)
        return 0;
        queue<pair<int, int>> q;
        q.push({sx, sy});
        int n = forest.size(), m = forest[0].size();
        vector<vector<int>> seen (n, vector<int> (m,0));
        seen[sx][sy]=1;
        int cost =0;
        while(!q.empty()) {
            int qsize = q.size();
            for(int i=0;i<qsize;i++) {
                pair<int, int> point = q.front();
                q.pop();
                for(int j=0;j<dir.size();j++) {
                    long long newx = point.first + dir[j][0];
                    long long newy = point.second + dir[j][1];
                    if(newx >=n || newy >=m ||newx<0|| newy<0)
                    continue;
                    if(forest[newx][newy]!=0 &&seen[newx][newy] ==0) {
                        q.push({newx, newy});
                        seen[newx][newy]=1;
                        if(newx == fx && newy ==fy) {
                            return cost+1;
                        }
                    }
                }
            }
            cost++;
        }
        return -1;

    }
};
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