<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 PathSize : public exception{
private:
    int m_pathSize;

public:
    PathSize(int pathSize): m_pathSize(pathSize){};
    const char* what() const noexcept override {
        return "Result found";
    }
    int getPathSize(){
        return m_pathSize;
    }
};

class Solution {
private:

    void updateQueue(int row, int col, const vector<vector<int>>& forest, int steps,
            queue<pair<int, pair<int,int>>>& stepsAndNeighbors, int destinationTree,
            vector<vector<int>>& isVisited){
        
        if(row <0 || row>= forest.size() || col<0 || col>= forest[row].size() || 
            isVisited[row][col])
            return;
        if(forest[row][col] == destinationTree){
            throw PathSize(steps);
        }
        if(forest[row][col] >=1){
            //cout<<"Insert neighbor in row = "<<row <<" col = "<<col<<" steps = "<<steps<<endl;
            stepsAndNeighbors.push(make_pair(steps, make_pair(row, col)));
        }
        isVisited[row][col] = true;


    }
    int countSteps(int& rowStart, int& colStart, int destinationTree, 
                   const vector<vector<int>>& forest){
        if(forest[rowStart][colStart]==destinationTree)
            return 0;
        vector<vector<int>> isVisited(forest.size(), vector<int> (forest[0].size(), false));
        queue<pair<int, pair<int,int>>> stepsAndNeighbors;
        stepsAndNeighbors.push(make_pair(0, make_pair(rowStart, colStart)));
        isVisited[rowStart][colStart] = true;

        while(!stepsAndNeighbors.empty()){
            
            auto p = stepsAndNeighbors.front();
            stepsAndNeighbors.pop();
            //cout<<"current vertex: row = "<<p.second.first <<" col = "<<p.second.second<<" steps = "<<p.first<<endl;
            try{
                //cout<<"new steps = "<<p.first +1<<endl;
                p.second.first++;
                updateQueue(p.second.first, p.second.second, forest, p.first +1,
                        stepsAndNeighbors, destinationTree,isVisited);
                p.second.first-=2;
                updateQueue(p.second.first, p.second.second, forest, p.first +1,
                            stepsAndNeighbors, destinationTree,isVisited);
                p.second.first++;
                p.second.second ++;
                updateQueue(p.second.first, p.second.second, forest, p.first +1,
                            stepsAndNeighbors, destinationTree,isVisited);
                p.second.second-=2;
                updateQueue(p.second.first, p.second.second, forest, p.first +1,
                            stepsAndNeighbors, destinationTree,isVisited);
            }
            catch(PathSize& pathSize){
                rowStart = p.second.first;
                colStart = p.second.second;
                return pathSize.getPathSize();
            }
            
            
        }
        return -1; 

    }
public:
    int cutOffTree(vector<vector<int>>& forest) {
        vector<int> trees;

        for(int i=0; i< forest.size(); i++){
            for(int j=0; j<forest[i].size(); j++){
                if(forest[i][j] >1){
                    trees.push_back(forest[i][j]);
                }
            }
        }

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

        int rowStart=0, colStart=0;
        int steps = 0;
        for(int tree: trees){
            //cout<<"current tree = "<<tree<<endl;
            int tempSteps = countSteps(rowStart, colStart, tree,forest);
            if(tempSteps == -1)
                return -1;
            steps+=tempSteps;
            //cout<<"After tree found steps = "<<steps<<endl<<endl;
        }
        return steps;
        
    }
};
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