11.12.2024刷华为OD-集合的使用,递归回溯的使用

HJ41 集合的使用

HJ43 迷宫问题–递归回溯的使用

def dfs(x, y, path, grid):n = len(grid)m = len(grid[0])if x == n-1 and y == m-1:for cor in path:print("({},{})".format(cor[0],cor[1]))# 判断条件：1不能越界 2不能撞墙 3走过的不能走if 0 <= x+1 < n and 0 <= y < m and grid[x+1][y] != 1:path.append((x+1, y))grid[x+1][y] = 1dfs(x+1, y, path, grid)path.pop()grid[x + 1][y] = 0if 0 <= x-1 < n and 0 <= y < m and grid[x-1][y] != 1:path.append((x-1, y))grid[x-1][y] = 1dfs(x-1, y, path, grid)path.pop()grid[x - 1][y] = 0if 0 <= x < n and 0 <= y+1 < m and grid[x][y+1] != 1:path.append((x, y+1))grid[x][y+1] = 1dfs(x, y+1, path, grid)path.pop()grid[x][y + 1] = 0if 0 <= x < n and 0 <= y-1 < m and grid[x][y-1] != 1:path.append((x, y-1))grid[x][y-1] = 1dfs(x, y-1, path, grid)path.pop()grid[x][y - 1] = 0if __name__ == "__main__":n, m = map(int, input().split())grid = []for _ in range(n):grid.append(list(map(int, input().split())))# print(grid)path = [(0, 0)]grid[0][0] = 1dfs(0,0,path, grid)

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