<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:



    long long cost(int i1, int j1, int i2, int j2, vector<vector<int>>& forest, int&n, int &m, vector<vector<int>>& dp, vector<vector<int>> &visited){        
        
        queue<pair<pair<int,int>,pair<int,int>>> q;
        q.push({{i1,j1},{i1,j1}});
        dp[i1][j1] = -1;
        visited[i1][j1] = 1;
        while(!q.empty()){
            auto a = q.front(); q.pop();

            int dx[4] = {1,0,-1,0};
            int dy[4] = {0,1,0,-1};

            dp[a.first.first][a.first.second] = dp[a.second.first][a.second.second]+1;

            for (int k=0;k<4;k++){
                int x = a.first.first+dx[k], y = a.first.second+dy[k];
                if(x < 0 || x >= n || y < 0 || y >= m || forest[x][y] == 0 || visited[x][y] == 1) continue;
                visited[x][y] = 1;
                q.push({{x,y},{a.first.first,a.first.second}});
            }
        }
        return dp[i2][j2];
        // if(dp[i1][j1] != -1){
        //     return dp[i1][j1];
        // }

        // if(i1 == i2 && j1 == j2){
        //     return dp[i1][j1] = 0;
        // }

        // int dx[4] = {1,0,-1,0};
        // int dy[4] = {0,1,0,-1};
        // long long ans = 1e8+1;
        // for(int k=0;k<4;k++){
        //     int x = i1+dx[k], y = j1+dy[k];
        //     if(x < 0 || x >= n || y < 0 || y >= m || forest[x][y] == 0) continue;
        //     visited[x][y] = 1;
        //     ans = min(ans, 1+cost(i1+dx[k], j1+dy[k],i2,j2,forest,n,m,dp,visited));
        // }
        // return dp[i1][j1] = ans;
    }

    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size();
        int m = forest[0].size();
        vector<pair<int,pair<int,int>>> treeCoord;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(forest[i][j] > 1){
                    treeCoord.push_back({forest[i][j],{i,j}});
                }
            }
        }
        sort(treeCoord.begin(), treeCoord.end());
        int numTree = treeCoord.size();
        vector<vector<int>> dp(n+1, vector<int>(m+1,-1)), visited(n+1, vector<int>(m+1,0));
        visited[0][0] = 1;
        int answer=cost(0,0,treeCoord[0].second.first, treeCoord[0].second.second, forest, n, m, dp, visited);
        if (answer == -1) return -1;
        dp.assign(n+1, vector<int>(m+1,-1));
        visited.assign(n+1, vector<int>(m+1,0));
        for(int i=0;i<numTree-1;i++){
            visited[treeCoord[i].second.first][treeCoord[i].second.second] = 1;
            int val = cost(treeCoord[i].second.first, treeCoord[i].second.second,treeCoord[i+1].second.first, treeCoord[i+1].second.second, forest, n, m, dp, visited);
            dp.assign(n+1, vector<int>(m+1,-1));
            visited.assign(n+1, vector<int>(m+1,0));
            if (val == -1) return -1;
            answer += val;
        }
        return answer;
    }
};
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