1. 火会扩散，但是我们可以看作火不会扩散到已经着火的格子，这样我们就可以记录每一个为草地的格子的着火时间
2. 在代码中，因为数字 2 已经表示墙了，所以我们把当时间为 0 时着火的格子在 gird 中的值设为 3，时间为 1 时着火的格子在 gird 中的值设为 4，以此类推，知道标注完所有可以着火的格子，通过BFS的方式就可以完成这一步
3. 通过 dfs(i, j, time) 的寻路方式来寻找可行的路径当 grid[i][j] <= time + 3 时说明 time 时刻 grid[i][j] 会着火，因此不能进行移动，除非它是终点

class Solution:
    def maximumMinutes(self, grid: List[List[int]]) -> int:
        fire = []
        m = len(grid)
        n = len(grid[0])
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    fire.append([i, j])
                    grid[i][j] = 3
        t = 4
        while len(fire) > 0:
            for _ in range(len(fire)):
                f = fire.pop(0)
                if f[0] + 1 < m and grid[f[0] + 1][f[1]] == 0:
                    fire.append([f[0] + 1, f[1]])
                    grid[f[0] + 1][f[1]] = t
                if f[0] - 1 >= 0 and grid[f[0] - 1][f[1]] == 0:
                    fire.append([f[0] - 1, f[1]])
                    grid[f[0] - 1][f[1]] = t
                if f[1] + 1 < n and grid[f[0]][f[1] + 1] == 0:
                    fire.append([f[0], f[1] + 1])
                    grid[f[0]][f[1] + 1] = t
                if f[1] - 1 >= 0 and grid[f[0]][f[1] - 1] == 0:
                    fire.append([f[0], f[1] - 1])
                    grid[f[0]][f[1] - 1] = t
            t += 1

        vis = [[0] * n for _ in range(m)]
        @cache
        def find(i, j, time):
            if grid[i][j] == 2 or (grid[i][j] != 0 and grid[i][j] <= time + 3):
                if i == m - 1 and j == n - 1 and grid[i][j] == time + 3:
                    return True 
                return False
            if i == m - 1 and j == n - 1:
                return True
            vis[i][j] = 1
            if i + 1 < m and vis[i + 1][j] == 0 and find(i + 1, j, time + 1)\
            or i > 0 and vis[i - 1][j] == 0 and find(i - 1, j, time + 1)\
            or j + 1 < n and vis[i][j + 1] == 0 and find(i, j + 1, time + 1)\
            or j > 0 and vis[i][j - 1] == 0 and find(i, j - 1, time + 1):
                return True
            vis[i][j] = 0
            return False

        if find(0, 0, t):
            return 10**9

        vis = [[0] * n for _ in range(m)]

        if find(0, 0, 0) is False:
            return -1

        l, r = 0, t - 3
        while l < r - 1:
            mid = (l + r) >> 1
            vis = [[0] * n for _ in range(m)]
            if find(0, 0, mid):
                l = mid
            else:
                r = mid
        
        return l