LeetCode 1293. Shortest Path in a Grid with Obstacles Elimination 下一层坑

最新推荐文章于 2021-09-26 11:29:38 发布

原创最新推荐文章于 2021-09-26 11:29:38 发布 · 390 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

475 篇文章

订阅专栏

leetcode

231 篇文章

订阅专栏

本文探讨了在一个充满障碍的网格中寻找从左上角到右下角的最短路径，允许消除一定数量的障碍。通过使用BFS算法，文章详细解释了如何实现这一目标，并提供了一个Python代码示例。

Given a m * n grid, where each cell is either 0 (empty) or 1 (obstacle). In one step, you can move up, down, left or right from and to an empty cell.

Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m-1, n-1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

Example 1:

Input: 
grid = 
[[0,0,0],
 [1,1,0],
 [0,0,0],
 [0,1,1],
 [0,0,0]], 
k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10. 
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

Example 2:

Input: 
grid = 
[[0,1,1],
 [1,1,1],
 [1,0,0]], 
k = 1
Output: -1
Explanation: 
We need to eliminate at least two obstacles to find such a walk.

Constraints:

grid.length == m
grid[0].length == n
1 <= m, n <= 40
1 <= k <= m*n
grid[i][j] == 0 or 1
grid[0][0] == grid[m-1][n-1] == 0

-----------------------------------------------------

BFS的时候下一轮条件刚开始没有想明白，写了个死循环的bug，题目并不难：

from collections import defaultdict

class Solution:
    def shortestPath(self, grid, k: int) -> int:
        m,n,step = len(grid),len(grid[0]),0
        layers,dic = [[(0,0,0)],[]],defaultdict(lambda:1600)
        dic[(0,0)] = 0
        cur,nxt = 0,1
        if (m == 1 and n == 1):
            return 0
        while (layers[cur] and step <= m*n):
            step += 1
            for x,y,ob in layers[cur]:
                for dx,dy in [[0,1],[0,-1],[-1,0],[1,0]]:
                    nx,ny = x+dx,y+dy
                    if (nx>=0 and nx<m and ny>=0 and ny<n):
                        nob = (ob+1) if grid[nx][ny] == 1 else ob
                        #bug1: if (nob<=k and (((nx,ny) not in dic) or dic[(nx,ny)]>=nob)): 而且只有((nx,ny) not in dic)才刷dic
                        if (nob<=k and dic[(nx,ny)]>nob):
                            if (nx == m-1 and ny == n-1):
                                return step
                            layers[nxt].append((nx,ny,nob))
                            dic[(nx,ny)] = nob
            layers[cur].clear()
            cur,nxt = nxt,cur
        return -1