<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:
    vector<int> dx = {-1, 1, 0, 0};
    vector<int> dy = {0, 0, 1, -1};
    int bfs(int srci,int srcj,int desti,int destj, int level, vector<vector<int>>& forest) {
        int ans = -1;
        
        queue<pair<int, pair<int,int>>> q;
        q.push({0, {srci, srcj}});

        vector<vector<bool>> vis(55, vector<bool>(55));

        while(!q.empty()){
            auto it = q.front();
            q.pop();

            int curri = it.second.first;
            int currj = it.second.second;
            int currlvl = it.first;

            if(curri==desti && currj==destj){
                return currlvl;
            }

            for(int i=0;i<4;i++){
                int newi = curri + dx[i];
                int newj = currj + dy[i];

                if(newi<forest.size() && newj<forest[0].size() && !vis[newi][newj] && forest[newi][newj]){
                    vis[newi][newj] = true;
                    q.push({currlvl+1, {newi, newj }});
                }
            }
        }

        return ans;

    }

    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size();
        int m = forest[0].size();

        vector<vector<bool>> visited(n, vector<bool>(m));

        vector<pair<int,pair<int,int>>> flatgrid;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(forest[i][j]>1){
                    flatgrid.push_back({forest[i][j], {i,j}});
                }
            }
        }

        flatgrid.push_back({0,{0,0}});

        sort(flatgrid.begin(), flatgrid.end());

        int elem = flatgrid.size(), fin = 0;

        for(int i=0;i<elem-1;i++){
            int srci = flatgrid[i].second.first;
            int srcj = flatgrid[i].second.second;

            int desti = flatgrid[i+1].second.first;
            int destj = flatgrid[i+1].second.second;

            int ans = bfs(srci, srcj, desti, destj, 0, forest);
            if(ans==-1){
                return -1;
            }
            fin += ans;
        }

        return fin;
    }
};
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