leetcode 2258. Escape the Spreading Fire（python）_王大呀呀

描述

You are given a 0-indexed 2D integer array grid of size m x n which represents a field. Each cell has one of three values:

0 represents grass,
1 represents fire,
2 represents a wall that you and fire cannot pass through.

You are situated in the top-left cell, (0, 0), and you want to travel to the safehouse at the bottom-right cell, (m - 1, n - 1). Every minute, you may move to an adjacent grass cell. After your move, every fire cell will spread to all adjacent cells that are not walls.

Return the maximum number of minutes that you can stay in your initial position before moving while still safely reaching the safehouse. If this is impossible, return -1. If you can always reach the safehouse regardless of the minutes stayed, return 109.

Note that even if the fire spreads to the safehouse immediately after you have reached it, it will be counted as safely reaching the safehouse.

A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).

Example 1:

Input: grid = [[0,2,0,0,0,0,0],[0,0,0,2,2,1,0],[0,2,0,0,1,2,0],[0,0,2,2,2,0,2],[0,0,0,0,0,0,0]]
Output: 3
Explanation: The figure above shows the scenario where you stay in the initial position for 3 minutes.
You will still be able to safely reach the safehouse.
Staying for more than 3 minutes will not allow you to safely reach the safehouse.

Example 2:

Input: grid = [[0,0,0,0],[0,1,2,0],[0,2,0,0]]
Output: -1
Explanation: The figure above shows the scenario where you immediately move towards the safehouse.
Fire will spread to any cell you move towards and it is impossible to safely reach the safehouse.
Thus, -1 is returned.

Example 3:

Input: grid = [[0,0,0],[2,2,0],[1,2,0]]
Output: 1000000000
Explanation: The figure above shows the initial grid.
Notice that the fire is contained by walls and you will always be able to safely reach the safehouse.
Thus, 109 is returned.

Note:

m == grid.length
n == grid[i].length
2 <= m, n <= 300
4 <= m * n <= 2 * 10^4
grid[i][j] is either 0, 1, or 2.
grid[0][0] == grid[m - 1][n - 1] == 0

解析

根據題意，給定一個大小為 m x n 的 0 索引 2D 整數陣列網格，每個單元格具有以下三個值之一：

0 代表草
1 代表火
2 代表不能穿過的牆

人一開始左上角的單元格 (0, 0) ，並且您想前往位於右下角的單元格 (m - 1, n - 1) 的安全屋。每分鐘都可以移動到相鄰的格子。人移動後的同時，每個火苗都會蔓延到所有相鄰的非牆壁單元。

返回移動前您可以在初始位置停留的最大分鐘數，同時仍能再移動後安全到達安全屋。如果這是不可能的，則返回 -1 。如果無論停留多長時間，您總能到達安全屋，請返回 10^9。請注意，即使在您到達後立即火災蔓延到安全屋，也將被視為安全到達安全屋。

這道題看起來比較嚇人，其實還比較簡單，因為要最少在初始位置停留為 0 分鐘，最多在初始位置停留 m*n 分鐘也就是 2 * 10^4 ，我們可以使用二分法來找出最少可以停留的時間，這是總體思路，然後我們可以使用 BFS 來計算出每個格子發生火災所需要的最短時間，這樣我們就能判斷人在某分鐘走到某格子會不會被燒焦，這樣不斷用二分法來找最少可以停留的時間即可。

時間複雜度為 O(MN) ，空間複雜度為 O(MN) 。

解答

class Solution(object):
    def maximumMinutes(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        M, N = len(grid), len(grid[0])
        timeArrive = [[float("inf") for _ in range(N)] for _ in range(M)]
        fire = []
        for i in range(M):
            for j in range(N):
                if grid[i][j] == 1:
                    fire.append([i, j])
                    timeArrive[i][j] = 0
        stack = []
        for i, j in fire:
            stack.append((i, j, 0))
        while stack:
            x, y, time = stack.pop(0)
            for i, j in [[x + 1, y], [x, y + 1], [x, y - 1], [x - 1, y]]:
                if 0 <= i < M and 0 <= j < N and grid[i][j] == 0 and timeArrive[i][j] > time + 1:
                    timeArrive[i][j] = time + 1
                    stack.append((i, j, time + 1))

        def waitK(k):
            h = [(0, 0, k)]
            vis = set()
            vis.add((0, 0))
            while h:
                x, y, time = h.pop(0)
                for i, j in [[x + 1, y], [x, y + 1], [x, y - 1], [x - 1, y]]:
                    if i == M - 1 and j == N - 1 and timeArrive[i][j] >= time + 1:
                        return True
                    if 0 <= i < M and 0 <= j < N and grid[i][j] == 0 and timeArrive[i][j] > time + 1 and (i, j) not in vis:
                        h.append((i, j, time + 1))
                        vis.add((i, j))
            return False

        if waitK(2*10**4):
            return 10**9
        elif not waitK(0):
            return -1
        l,r = 0, 2*10**4
        while l<r:
            mid = (l+r)//2
            if waitK(mid):
                if not waitK(mid+1):
                    return mid
                else:
                    l = mid + 1
            else:
                r = mid
        return l

執行結果

55 / 55 test cases passed.
Status: Accepted
Runtime: 1033 ms
Memory Usage: 16.2 MB

原題連結

http://leetcode.com/contest/biweekly-contest-77/problems/escape-the-spreading-fire/

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