<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 cutOffTree(vector<vector<int>>& forest) {
        vector<tuple<int, int, int>> location{{0, 0, 0}};
        for (int i = 0; i < forest.size(); i++) {
            for (int j = 0; j < forest[0].size(); j++) {
                if (forest[i][j] > 1) {
                    location.emplace_back(forest[i][j], i, j);
                }
            }
        }
        sort(location.begin(), location.end());

        auto bfs = [&](int sx, int sy, int ex, int ey) {
            if (sx == ex && sy == ey) {
                return 0;
            }
            vector<vector<bool>> visited(forest.size(), vector<bool>(forest[0].size()));
            queue<tuple<int, int, int>> que;
            visited[sx][sy] = true;
            que.push({0, sx, sy});

            const vector<int> dx{1, 0, -1, 0};
            const vector<int> dy{0, 1, 0, -1};

            while (!que.empty()) {
                auto [step, x, y] = que.front();
                que.pop();
                for (int i = 0; i < dx.size(); i++) {
                    int nx = x + dx[i];
                    int ny = y + dy[i];
                    if (0 <= nx && nx < forest.size() && 0 <= ny && ny < forest[0].size() &&  forest[nx][ny] && !visited[nx][ny]) {
                        if (nx == ex && ny == ey) {
                            return step + 1;
                        }
                        visited[nx][ny] = true;
                        que.push({step + 1, nx, ny});
                    }
                }
            }

            return -1;
        };

        int total_distance = 0;
        for (int i = 0; i < location.size() - 1; i++) {
            auto [_, sx, sy] = location[i];
            auto [__, ex, ey] = location[i + 1];
            int distance = bfs(sx, sy, ex, ey);
            if (distance == -1) {
                total_distance = -1;
                break;
            }
            total_distance += bfs(sx, sy, ex, ey);
        }
        return total_distance;
    }
};
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