<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 {
private:
    int minCostWithGoal(vector<vector<int>>& forest, int &cx, int &cy, int goal){
        int ans = -1;
        vector<int> dx = {0,1,0,-1};
        vector<int> dy = {1,0,-1,0};
        unordered_set<int> seen; // hash via 50*,0
        queue<pair<int,int>> bfs;
        seen.insert(50* cx+ cy);
        int cpx = cx;
        int cpy = cy;
        bfs.push({cpx,cpy});
        while(!bfs.empty()){
            int layerSZ = bfs.size();
            ++ans;
            while(layerSZ){
                --layerSZ;
                int x = bfs.front().first;
                int y = bfs.front().second;
                bfs.pop();
                // cout << " exp " << x << "," << y << " || ";
                if(forest[x][y] == goal){
                    cx = x;
                    cy = y;
                    // cout << " found " << goal << endl;
                    return ans;
                }
                else{
                    for(int i = 0; i < 4; ++i){
                        int nx = x + dx[i];
                        int ny = y + dy[i];
                        if(
                            nx > -1 && nx < forest.size()
                            && ny > -1 && ny < forest[0].size()
                            && !seen.count(50*nx + ny) && forest[nx][ny] != 0 
                        ){
                            bfs.push({nx,ny});
                            seen.insert(50*nx + ny);
                        }
                    }
                }
            }
        }
        return -1;
    }
public:
    int cutOffTree(vector<vector<int>>& forest) {
        // 
        vector<int> uniqueElements;
        for(auto v : forest){
            for(int i: v){
                if(i > 1) uniqueElements.push_back(i);
            }
        }
        sort(uniqueElements.begin(), uniqueElements.end());
        int ans = 0, cx = 0, cy = 0;
        for(int i = 0; i < uniqueElements.size(); ++i){
            // cout << uniqueElements[i] << " -> ";
            int stepNext = minCostWithGoal(forest, cx, cy, uniqueElements[i]);
            // cout << cx << " : " << cy << endl; 
            if(stepNext == -1) return -1;
            ans += stepNext;
        }
        return ans;
    }
};
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