Leetcode 3552. Grid Teleportation Traversal

Espresso Macchiato2025-05-19 21:25

[Leetcode 3552. Grid Teleportation Traversal](#Leetcode 3552. Grid Teleportation Traversal)
- [1. 解题思路](#1. 解题思路)
- [2. 代码实现](#2. 代码实现)

题目链接：3552. Grid Teleportation Traversal

1. 解题思路

这一题的话核心就是一个广度优先遍历，我们只需要从原点开始，一点点考察其所能到达的位置，直至其最终到达终点即可。

唯一特殊的是，考虑到传送的存在，这里会有些特殊操作，不难发现，一个端口至多发生一次传送，否则必然可以优化路线，因此我们只需要考察每一个端口值上是否有过传送即可。

2. 代码实现

给出python代码实现如下：

python 复制代码

class Solution:
    def minMoves(self, matrix: List[str]) -> int:
        n, m = len(matrix), len(matrix[0])
        teleportation = defaultdict(list)
        for i in range(n):
            for j in range(m):
                if matrix[i][j] in "#.":
                    continue
                teleportation[matrix[i][j]].append((i, j))
                
        seen = {(0, 0)}
        q = [(0, 0, 0, 0)]
        
        def add_teleportation(step, i, j, status):
            nonlocal seen, q
            if matrix[i][j] not in "#." and (status & (1 << (ord(matrix[i][j]) - ord('A'))) == 0):
                for ti, tj in teleportation[matrix[i][j]]:
                    if (ti, tj) not in seen:
                        heapq.heappush(q, (step, ti, tj, status | (1 << (ord(matrix[i][j]) - ord('A')))))
                        seen.add((ti, tj))
            return
        
        add_teleportation(0, 0, 0, 0)
        
        while q != []:
            step, i, j, status = heapq.heappop(q)
            if (i, j) == (n-1, m-1):
                return step
            if i+1 < n and (i+1, j) not in seen and matrix[i+1][j] != "#":
                heapq.heappush(q, (step+1, i+1, j, status))
                seen.add((i+1, j))
                add_teleportation(step+1, i+1, j, status)
            if j+1 < m and (i, j+1 ) not in seen and matrix[i][j+1] != "#":
                heapq.heappush(q, (step+1, i, j+1, status))
                seen.add((i, j+1))
                add_teleportation(step+1, i, j+1, status)
            if i-1 >= 0 and (i-1, j) not in seen and matrix[i-1][j] != "#":
                heapq.heappush(q, (step+1, i-1, j, status))
                seen.add((i-1, j))
                add_teleportation(step+1, i-1, j, status)
            if j-1 >= 0 and (i, j-1 ) not in seen and matrix[i][j-1] != "#":
                heapq.heappush(q, (step+1, i, j-1, status))
                seen.add((i, j-1))
                add_teleportation(step+1, i, j-1, status)
        return -1

提交代码评测得到：耗时6943ms，占用内存147.7MB。