<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>

#define mypair pair<int,pair<int,int>>
class Solution {
public:
int check(int x,int y,vector<vector<int>>& forest){
    int n=forest.size();
    int m=forest[0].size();
    if(x>=0 && x<n && y>=0 && y<m && forest[x][y] >= 1) return 1;
    return 0;
}
int bfs(vector<vector<int>>& forest,int sx,int sy,int ex,int ey){
    if(sx==ex && sy==ey) return 0;
    int n=forest.size();
    int m=forest[0].size();
    vector<vector<int>> vis(n,vector<int>(m,0));
    queue<mypair> q;
    q.push({0,{sx,sy}});
    vis[sx][sy]=1;
    int dx[4]={0,1,0,-1};
    int dy[4]={1,0,-1,0};
    while(!q.empty()){
        mypair node=q.front();
        q.pop();
        sx=node.second.first,sy=node.second.second;
        for(int k=0;k<4;k++){
            int nx=sx+dx[k];
            int ny=sy+dy[k];
            if(check(nx,ny,forest)){
                if(!vis[nx][ny]){
                    vis[nx][ny]=1;
                    if(nx==ex && ny==ey){
                        forest[nx][ny]=1;
                        return node.first+1;
                    }
                    q.push({node.first+1,{nx,ny}});
                }
                
            }
        }
        
    }
    return -1;
}
    int cutOffTree(vector<vector<int>>& forest) {
        int n=forest.size();
        int m=forest[0].size();
        vector<mypair> trees;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if (forest[i][j] > 1) {
                trees.push_back({forest[i][j],{i,j}});
                }
            }
        }
        sort(trees.begin(),trees.end());
        int step=0;
        int sx=0,sy=0;
        for(int i=0;i<trees.size();i++){
            int ex=trees[i].second.first;
            int ey=trees[i].second.second;
            int req=bfs(forest,sx,sy,ex,ey);
            if(req<0) return -1;
            step+=req;
            sx=ex;
            sy=ey;

        }
        return step;

    }
};
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