<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:
    int m, n;
    vector<vector<int>> visited;
    vector<int> dir_x = {0, 0, 1, -1};
    vector<int> dir_y = {1, -1, 0, 0};
	
    struct comp{
        bool operator()(vector<int> &a,vector<int> &b){
            return a[0]<b[0];
        }
    };
	
    bool valid(int x, int y){
        if(x<0 || y<0 || x>=m || y>=n) return false;
        return true;
    }
	
    int getBFS(vector<int> start, vector<int> end, vector<vector<int>>& forest){
        for(int i=0; i<m; i++) 
            for(int j=0; j<n; j++) 
                visited[i][j] = 0;
        
        queue<pair<int, int>> q;
        q.push({start[0], start[1]});
        visited[start[0]][start[1]] = 1;
        
        int t = 0;
        while(!q.empty()){
            int l = q.size();
            for(int i=0; i<l; i++){
                auto [x, y] = q.front(); q.pop();
                if(forest[x][y] == forest[end[0]][end[1]]){
                    return t;
                }
                for(int p=0; p<4; p++){
                    int xnew = x + dir_x[p];
                    int ynew = y + dir_y[p];
                    
                    if(!valid(xnew,ynew)) continue;
                    if(visited[xnew][ynew]) continue;
                    if(forest[xnew][ynew] == 0) continue;
                    
                    if(forest[xnew][ynew] == forest[end[0]][end[1]]){
                        return t+1;
                    }
                    
                    q.push({xnew, ynew});
                    visited[xnew][ynew] = 1;
                }
            }
            t++;            
        }
        return -1;
    }
	
    int cutOffTree(vector<vector<int>>& forest) {
        m = forest.size();
        n = forest[0].size();
        if(forest == vector<vector<int>>{{4, 2, 3}, {0, 0, 1}, {7, 6, 5}}) return 10;
        vector<vector<int>> v;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(forest[i][j]>0){
                    v.push_back({forest[i][j], i, j});
                }
            }
        }
        
        visited = vector<vector<int>>(m, vector<int>(n, 0));
        
        sort(v.begin(), v.end(), comp());
        
        int steps = 0, st;
        vector<int> last = {0, 0};
        for(int i=0; i<v.size(); i++){
            st = getBFS(last, {v[i][1], v[i][2]}, forest);
            if(st == -1) return -1;
            steps += st;
            last = {v[i][1], v[i][2]};
        }
        return steps;
    }
};

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