<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:
    typedef struct Node {
        int val, i, j;
    } Node;

    typedef struct AStarNode {
        int f, g, i, j;  // f(node) = g(node, src) + h(node, dst)
    } AStarNode;

    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size(), n = forest[0].size();
        vector<Node> trees;
        for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            if (forest[i][j] > 1)
                trees.push_back({forest[i][j], i, j});
        sort(trees.begin(), trees.end(), [](const Node& a, const Node& b) -> bool {
            return a.val < b.val;
        });
        int ans = 0, sr = 0, sc = 0;
        for (const auto& tree : trees) {
            auto [height, tr, tc] = tree;
            int d = dist(forest, sr, sc, tr, tc);
            if (d < 0) return -1;
            ans += d;
            // forest[tr][tc] = 1;  // cut off the tree
            sr = tr, sc = tc;
        }
        return ans;
    }

    int dist(vector<vector<int>>& forest, int sr, int sc, int tr, int tc) {
        // return bfs(forest, sr, sc, tr, tc);
        return astar(forest, sr, sc, tr, tc);
    }

    int bfs(vector<vector<int>>& forest, int sr, int sc, int tr, int tc) {
        int m = forest.size(), n = forest[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        vector<vector<int>> dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        queue<Node> q;
        q.push({0, sr, sc});
        while (!q.empty()) {
            auto [d, i, j] = q.front(); q.pop();
            if (i == tr && j == tc) return d;
            for (const auto& dir : dirs) {
                int next_i = i + dir[0];
                int next_j = j + dir[1];
                if (next_i < 0 || next_i >= m || next_j < 0 || next_j >= n || 
                    forest[next_i][next_j] == 0 || visited[next_i][next_j])
                    continue;
                q.push({d + 1, next_i, next_j});
                visited[next_i][next_j] = true;
            }
        }
        return -1;
    }

    int astar(vector<vector<int>>& forest, int sr, int sc, int tr, int tc) {
        int m = forest.size(), n = forest[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        vector<vector<int>> cost(m, vector<int>(n, INT_MAX));
        vector<vector<int>> dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        auto cmp = [](const AStarNode& a, const AStarNode& b) -> bool {
            return a.f > b.f;
        };
        priority_queue<AStarNode, vector<AStarNode>, decltype(cmp)> minHeap(cmp);
        minHeap.push({0, 0, sr, sc});
        cost[sr][sc] = 0;
        while (!minHeap.empty()) {
            auto [f, g, i, j] = minHeap.top(); minHeap.pop();
            if (i == tr && j == tc) return g;
            visited[sr][sc] = true;
            for (const auto& dir : dirs) {
                int next_i = i + dir[0];
                int next_j = j + dir[1];
                if (next_i < 0 || next_i >= m || next_j < 0 || next_j >= n || 
                    forest[next_i][next_j] == 0 || visited[next_i][next_j])
                    continue;
                //   dist(i,j,next_i,next_j)
                // = g(i,j,next_i,next_j) + h(next_i,next_j)
                // = 1 + abs(next_i - tr) + abs(next_j - tc)
                int relaxedCost = g + 1 + abs(next_i - tr) + abs(next_j - tc);
                if (cost[next_i][next_j] > relaxedCost) {
                    cost[next_i][next_j] = relaxedCost;
                    minHeap.push({relaxedCost, g + 1, next_i, next_j});
                }
            }
        }
        return -1;
    }
};
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