一个游戏为贪吃蛇，另一个游戏maze
在string里面能够直接看到迷宫
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000000000000000000000
0 0 0 0 0 0 0
0 0 0 00000 00000 0 0
0 0 0 0
0 000 000 0 000 0 0 0
0 0 0 0 0 0 0 0
0 0 0 00000 000 000 0
0 0 0 0 0 0
0 000 0 0 000 0 0 0 0
0 0 0 0 0 0 0 0 0
0 00000 000 000 0 0 0
0 0 0 0 0
000 0 0 0 000 0 0 0 0
0 0 0 0 0 0 * 0 0 0 0
0 0000000 0 000 00000
@ 0 0 0
0 0 0 0 0 00000000000
0 0 0 0 0
000 0 00000 0 000 000
0 0 0 0 0
000000000000000000000
虽然迷宫的地图已经出来了，但是下面还有一个v14[4] = '0';
所以真实的地图应该为
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000000000000000000000
0 0 0 0 0 0 0
0 0 0 00000 00000 0 0
0 0 0 0
0 000 000 0 000 0 0 0
0 0 0 0 0 0 0 0
0 0 0 00000 000 000 0
0 0 0 0 0 0
0 000 0 0 000 0 0 0 0
0 0 0 0 0 0 0 0 0
0 00000 000 000 0 0 0
0 0 0 0 0
000 0 0 0 000 0 0 0 0
0 0 0 0 0 0 * 0 0 0 0
0 0000000 0 000 00000
@ 0 0 0 0
0 0 0 0 0 00000000000
0 0 0 0 0
000 0 00000 0 000 000
0 0 0 0 0
000000000000000000000
从@走到*救出小猫，然后走到出口
路径脚本
点击查看代码
from collections import deque# 设置二维四向迷宫, 如果题目是多个小迷宫问题, 拆分多次调用脚本获取路径即可
# 每一排后面要逗号隔开，对齐++++++++++++++++++++++++++
maze = [[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],[1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1],[1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1],[1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1],[1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1],[1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1],[1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1],[1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0],[1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1],[1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1],[1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1],[1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1],[1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1],[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1],
[1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1],
[0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1],
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1],
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]
path_len = 0x7fffffff # 如果题目未给出终点坐标，则一定会指定路径的长度，在此处修改路径长度，否则请保留path_len的极大值
#已知步长时更改，否则不管+++++++++++++++++++++++++++++++++# 进行BFS寻找路径
def bfs(start, end, barrier):directions = [(0, 1), (1, 0), (0, -1), (-1, 0)] # 定义四个方向的移动for i in range(len(maze)): # 获取起点和终点在列表中的索引for j in range(len(maze[i])):if (maze[i][j] == start):start = (i, j)if (maze[i][j] == end):end = (i, j)# 以下均是bfs算法套路queue = deque()queue.append((start, [start])) # (当前位置, 路径)visited = set()visited.add(start)while queue:position, path = queue.popleft()if position == end:return pathelif len(path) == path_len:return pathfor d in directions:next_position = (position[0] + d[0], position[1] + d[1])if 0 <= next_position[0] < len(maze) and 0 <= next_position[1] < len(maze[0]) and \maze[next_position[0]][next_position[1]] != barrier and next_position not in visited:queue.append((next_position, path + [next_position]))visited.add(next_position)return None# 执行BFS搜索并打印结果
if __name__ == '__main__':#更改起点和终点的坐标，如果知道path而坐标未知则注释掉+++++++++++++++++++++++++++++maze[15][0] = 'S' #如果题目给了起点终点的坐标，在这里直接给起点和终点添加特征maze[7][20] = 'E'#前为横坐标，后为纵坐标path = bfs('S', 'E', 1) # bfs函数传入参数代表起点、终点、障碍的特征(若题目给出的数据无特征, 手动添加特征即可, 通常障碍是1也有可能是0或其它字符如'#')print("移动路径坐标：", path)print("移动路径方位：{", end='')for i in range(1, len(path)):x1, y1, x2, y2 = path[i - 1][0], path[i - 1][1], path[i][0], path[i][1]if (x1 > x2): # 上print("w", end='')elif (x1 < x2): # 下print("s", end='')elif (y1 > y2): # 左print("a", end='')elif (y1 < y2): # 右print("d", end='')print('}')
maze下面有score的检验
异或
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print(0x1DC4 ^ 13376013)
#13371337
flag格式为
HZCTF{md5(path)+score}
NSSCTF
