Leetcode刷题笔记-两点间距离

675. Cut Off Trees for Golf Event

class Solution(object):
    def cutOffTree(self, forest):

        # add border 0 so we do not need to index-checks later on
        forest.append([0] * len(forest[0]))
        for row in forest:
            row.append(0)
            
        # find the trees:
        trees = [(height, i, j) for i, row in enumerate(forest)
                 for j, height in enumerate(row) if height > 1]
        
        # Can we reach all the trees? if not return -1 right away
        queue = [(0, 0)]
        reached = set()  # all the i,j can be reached
        for i, j in queue:
            if i >= 0 and j >= 0 and (i,j) not in reached and forest[i][j]:
                reached.add((i, j))
                queue += (i+1, j), (i-1, j), (i, j+1), (i, j-1)
               
        if not all((i,j) in reached for (_, i, j) in trees):
            return -1
        
        # Distance from (i, j) to (I, J).
        def distance(i, j, I, J):
            now, soon = [(i, j)], []
            expended = set()  # 已经走过的
            manhatten = abs(i-I) + abs(j-J)
            detours = 0
            while True:
                if not now:
                    now, soon = soon, []
                    detours += 1
                i, j = now.pop()
                if (i, j) == (I, J):
                    return manhatten + 2*detours
                
                if (i, j) not in expended:
                    expended.add((i, j))
                    for i, j, closer in ((i+1, j, i < I), (i-1, j, i > I), (i, j+1, j<J), (i, j-1, j>J)):
                        if forest[i][j]:
                            (now if closer else soon).append((i, j))
        trees.sort()
        return sum(distance(i, j, I, J) for (_, i, j), (_, I, J) in zip([(0, 0, 0)] + trees, trees))