You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map, in this map:

1. 0 represents the obstacle can‘t be reached.
2. 1 represents the ground can be walked through.
3. The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the tree‘s height.

You are asked to cut off all the trees in this forest in the order of tree‘s height - always cut off the tree with lowest height first. And after cutting, the original place has the tree will become a grass (value 1).

You will start from the point (0, 0) and you should output the minimum steps you need to walk

You are guaranteed that no two trees have the same height and there is at least one tree needs to be cut off.

Example 1:

Input: 
[
 [1,2,3],
 [0,0,4],
 [7,6,5]
]
Output: 6

Example 2:

Input: 
[
 [1,2,3],
 [0,0,0],
 [7,6,5]
]
Output: -1

Example 3:

Input: 
[
 [2,3,4],
 [0,0,5],
 [8,7,6]
]
Output: 6
Explanation: You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking. 

Hint: size of the given matrix will not exceed 50x50.

為一個高爾夫賽事砍掉森林中所有高度大於1的樹，要按從低到高的順序砍。森林用一個2D的map來表示，0代表障礙物，無法通過。1代表地面，可以通過。其他整數代表是樹和相應的高度，可以通過。

解法：把是樹的節點，按樹高從低到高排序。然後從第一棵樹開始，每次都用BFS求出和下一棵樹之間的最短路徑，然後累計路徑和為結果。如果不能走到下一棵樹，則返回-1。

Python:

class Solution(object):
    def cutOffTree(self, forest):
        """
        :type forest: List[List[int]]
        :rtype: int
        """
        def dot(p1, p2):
            return p1[0]*p2[0]+p1[1]*p2[1]

        def minStep(p1, p2):
            min_steps = abs(p1[0]-p2[0])+abs(p1[1]-p2[1])
            closer, detour = [p1], []
            lookup = set()
            while True:
                if not closer:  # cannot find a path in the closer expansions
                    if not detour:  # no other possible path
                        return -1
                    # try other possible paths in detour expansions with extra 2-step cost
                    min_steps += 2
                    closer, detour = detour, closer
                i, j = closer.pop()
                if (i, j) == p2:
                    return min_steps
                if (i, j) not in lookup:
                    lookup.add((i, j))
                    for I, J in (i+1, j), (i-1, j), (i, j+1), (i, j-1):
                        if 0 <= I < m and 0 <= J < n and forest[I][J] and (I, J) not in lookup:
                            is_closer = dot((I-i, J-j), (p2[0]-i, p2[1]-j)) > 0
                            (closer if is_closer else detour).append((I, J))
            return min_steps

        m, n = len(forest), len(forest[0])
        min_heap = []
        for i in xrange(m):
            for j in xrange(n):
                if forest[i][j] > 1:
                    heapq.heappush(min_heap, (forest[i][j], (i, j)))

        start = (0, 0)
        result = 0
        while min_heap:
            tree = heapq.heappop(min_heap)
            step = minStep(start, tree[1])
            if step < 0:
                return -1
            result += step
            start = tree[1]
        return result　　

C++:

class Solution {
public:
    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size(), n = forest[0].size(), res = 0, row = 0, col = 0;
        vector<vector<int>> 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());
        for (int i = 0; i < trees.size(); ++i) {
            int cnt = helper(forest, row, col, trees[i][1], trees[i][2]);
            if (cnt == -1) return -1;
            res += cnt;
            row = trees[i][1];
            col = trees[i][2];
        }
        return res;
    }
    int helper(vector<vector<int>>& forest, int row, int col, int treeRow, int treeCol) {
        if (row == treeRow && col == treeCol) return 0;
        int m = forest.size(), n = forest[0].size(), cnt = 0;
        queue<pair<int, int>> q{{{row, col}}};
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        vector<vector<int>> dirs{{-1,0},{0,1},{1,0},{0,-1}};
        while (!q.empty()) {
            ++cnt;
            for (int i = q.size() - 1; i >= 0; --i) {
                auto t = q.front(); q.pop();
                for (auto dir : dirs) {
                    int x = t.first + dir[0], y = t.second + dir[1];
                    if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y] || forest[x][y] == 0) continue;
                    if (x == treeRow && y == treeCol) return cnt;
                    visited[x][y] = true;
                    q.push({x, y});
                }
            }
        }
        return -1;
    }
};

[LeetCode] 675. Cut Off Trees for Golf Event 為高爾夫賽事砍樹