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:
0represents theobstaclecan't be reached.1represents thegroundcan be walked through.The place with number bigger than 1represents atreecan 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 to cut off all the trees. If you can't cut off all the trees, output -1 in that situation.
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.
为一个高尔夫赛事砍掉森林中所有高度大于1的树，要按从低到高的顺序砍。森林用一个2D的map来表示，0代表障碍物，无法通过。1代表地面，可以通过。其他整数代表是树和相应的高度，可以通过。
解法：把是树的节点，按树高从低到高排序。然后从第一棵树开始，每次都用BFS求出和下一棵树之间的最短路径，然后累计路径和为结果。如果不能走到下一棵树，则返回-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;
}
};