<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> bfs(int src, int des, vector<vector<int>>& graph){
        int found = 0;
        int steps = 0;
        queue<int> q;
        q.push(src);
        vector<bool> vis(graph.size(),false);
        while(q.size()!=0){
            steps++;
            int s = q.size();
            for(int i = 1; i<=s; i++){
                int x = q.front(); q.pop();
                if(vis[x])continue;
                vis[x] = true;
            for(auto& it : graph[x]){
                if(it==des){
                    found = true; break;
                }
                if(vis[it]==false){
                    q.push(it);
                }
            }
            if(found)break;
            }
            if(found)break;
        }
        return vector<int>{steps,found};
    }
    int cutOffTree(vector<vector<int>>& forest) {
        vector<vector<int>> graph(forest.size()*forest[0].size()+2);
        vector<pair<int,int>> trees;
        if(forest[0][0]==0)return -1;
        for(int i = 0; i<forest.size(); i++){
        for(int j = 0; j<forest[0].size(); j++){
            if(j+1<forest[0].size() && forest[i][j] >0 && forest[i][j+1]>0){
                graph[i*forest[0].size() + j].push_back(i*forest[0].size() + j + 1);
                graph[i*forest[0].size() + j+1].push_back(i*forest[0].size() + j);

            }
            if(i-1>-1 && forest[i][j] >0 && forest[i-1][j]>0){
                graph[i*forest[0].size() + j].push_back((i-1)*forest[0].size() + j);
                graph[(i-1)*forest[0].size() + j].push_back(i*forest[0].size() + j);

            }
        

    if(forest[i][j]>1){
            trees.push_back({forest[i][j],i*forest[0].size() + j});
        }}
        }
        sort(trees.begin(),trees.end());
        int src = 0;
        int steps = 0;
        for(int i = 0; i<trees.size(); i++){
            if(trees[i].second==src){continue;}
            auto x = bfs(src,trees[i].second,graph);
            if(x[1]==0)return -1;
            steps+=x[0];
            src = trees[i].second;
        }
        return steps;
    }
};
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