動機

最小路徑? no,no 最短路徑!!!

Problem

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

 

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]Output: 2Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]Output: 1Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]Output: 0Explanation: This route does not require any effort.

 

Constraints:

  • rows == heights.length
  • columns == heights[i].length
  • 1 <= rows, columns <= 100
  • 1 <= heights[i][j] <= 106

Sol

from sortedcontainers import SortedList
class Solution:
    def minimumEffortPath(self, hs: List[List[int]]) -> int:
        legal = lambda i,j: 0 <= i < len(hs) and 0 <= j < len(hs[0])
        hq = [(0,(0,0))]
        efferts = defaultdict(lambda : float('inf'))
        
        while hq:
            cur, (x,y) = heappop(hq)
            if x == len(hs)-1 and y == len(hs[x])-1:
                return cur
            else:
                dirs = [(x+1,y),(x,y+1),(x-1,y),(x,y-1)]
                dirs = [(a,b) for a,b in dirs if legal(a,b)]
                for a,b in dirs:
                    now = max(cur,abs(hs[a][b]-hs[x][y]))
                    if now < efferts[(a,b)]:
                        efferts[(a,b)] = now
                        heappush(hq, (efferts[(a,b)], (a,b)))
        return -1