<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 {
private:
    vector<vector<int>> dirs = {
        {-1, 0}, {0, -1}, {0, 1}, {1, 0}
    };
    int hash(int i, int j, int m) {
        return i*m + j;
    }


public:
    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size(), m = forest[0].size();
        vector<pair<int, int>> ps;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int el = forest[i][j];
                if (max(i, j) == 0) {
                    ps.push_back(make_pair(-1, this->hash(i, j, m)));
                }
                if (el > 1) {
                    ps.push_back(make_pair(el, this->hash(i, j, m)));
                }
            }
        }

        sort(ps.begin(), ps.end());
        int pn = ps.size(), ret = 0;
        for (int pi = 0; pi < pn-1; pi++) {
            int curHash = ps[pi].second;
            int targetHash = ps[pi+1].second;

            vector<bool> seenHashes(n*m, false);
            queue<pair<int, int>> q;
            q.push({curHash, 0});
            int retDist = -1;
            while (q.size() > 0) {
                int h = q.front().first, dist = q.front().second;
                q.pop();

                if (h == targetHash) {
                    retDist = dist;
                    break;
                }

                if (!seenHashes[h]) {
                    seenHashes[h] = true;
                    int ci = h / m, cj = h % m;
                    for (auto& v: this->dirs) {
                        int ni = ci+v[0], nj = cj+v[1];
                        if (0 <= min(ni, nj) and ni < n and nj < m and 
                            forest[ni][nj] > 0) {
                            q.push(make_pair(this->hash(ni, nj, m), dist + 1));
                        }
                    }
                }
            }

            if (retDist == -1) {
                return -1;
            }
            ret += retDist;
        }
        return ret;
    }
};
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