<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 build(vector<vector<int>>& forest, int x1, int y1, int x2, int y2) {
        vector<pair<int, int> > q;
        q.push_back({x1, y1});
        int head = 0, tail = 1;
        int n = forest.size(), m = forest[0].size();
        vector<vector<int> > d(n, vector<int> (m, 1e9));
        d[x1][y1] = 0;
        while (head < tail) {
            pair<int, int> cur = q[head];
            if (cur.first == x2 && cur.second == y2) {
                return d[cur.first][cur.second];
            }
            int x = cur.first, y = cur.second;
            for (int dx = -1; dx <= 1; ++dx) {
                for (int dy = -1; dy <= 1; ++dy) {
                    if (dx * dy != 0) continue;
                    if (dx == 0 && dy == 0) continue;
                    int nx = x + dx, ny = y + dy;
                    if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
                    if (forest[nx][ny] == 0) continue;
                    if (d[nx][ny] > d[x][y] + 1) {
                        d[nx][ny] = d[x][y] + 1;
                        q.push_back({nx, ny});
                        ++tail;
                    }
                }
            }
            ++head;
        }
        return -1;
    }

    int cutOffTree(vector<vector<int>>& forest) {
        if (forest[0][0] == 0) return -1;
        int ans = 0;        
        vector<vector<int> > ids;
        ids.push_back({0, 0, 0});
        for (int i = 0; i < forest.size(); ++i) {
            for (int j = 0; j < forest[i].size(); ++j) {
                if (forest[i][j] >= 2) ids.push_back({forest[i][j], i, j});
            }
        }    
        sort(ids.begin(), ids.end());
        for (int i = 0; i + 1 < ids.size(); ++i) {
            int path = build(forest, ids[i][1], ids[i][2], ids[i + 1][1], ids[i + 1][2]);
            if (path == -1) return -1;
            ans += path;
            forest[ids[i][1]][ids[i][2]] = 1;
        }
        return ans;
    }
};
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