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re-二维四向迷宫题

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   约 5905 字   预计阅读 12 分钟 
https://bing.ee123.net/img/rand?artid=146264038

re-二维四向迷宫题

关于迷宫

我们结合具体的来进行描述。

迷宫，顾名思义是有墙，且只有一条路可以走下去。当我们在题目中获得了map（地图）就拥有了上帝视角，就可以顺利的走出迷宫。

在下面这个图就是一个迷宫的map，其中A是起点，B是终点，1代表墙，0代表通路（可以行走的区域）

https://i-blog.csdnimg.cn/direct/71fd03bac9cd4053a346529bb89e82c0.png 这仅仅只是一个88的地图，可以直接解决找到路径，但是如果遇到3030或者更大的地图该如何办？

迷宫的解决不可能完全依靠我们人力去走通，而要尽力依靠脚本。

写脚本之前我们得先了解一下在这个迷宫题里的上下左右的方向分别是什么

eg.大部分游戏中的上下左右就是WSAD

通常这个走的路径就是我们需要的关键信息（甚至可能就是flag）

比如在这个地图里我们的关键信息就是SDSDDSDSSD（假设方向同上）

而写脚本的好处就是不那么废眼睛，要不一旦遇到稍大的地图将会累眼睛也废精力且费时。

解题

有了部分关于迷宫的知识后，我们去学习一下如何解决，这里与实题结合

得到附件确定了是多少位的文件以及有无壳后（有壳得脱壳）用ida打开看

提取迷宫地图

进入后，shift+f12查看字符获得地图的位置，然后双击其中一行去到IDA View-A中进行提取数据

第一种类型（字符型地图）

如果是字符型的map就选中地图后按shift+e进行提取数据。这个除了提取数据，提取后还得辨别那个字符代表路径，哪个字符代表墙

f1

如果可以直接获取字符就选择string literal但是 https://i-blog.csdnimg.cn/direct/64e06d04490a4269934c1403c7a59a2c.png

f2

但是可能提取不了所有的，比如这个地图，那就选择另一个C unsigned char array(hex)，也就是将字符转为十六进制进行全部提取，然后再写脚本转换为地图 https://i-blog.csdnimg.cn/direct/2f23a996108f484087f789493c33f7c1.png

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  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x24, 
  0x24, 0x24, 0x24, 0x2E, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x2E, 0x00, 0x2E, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 
  0x24, 0x24, 0x24, 0x24, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x24, 0x2E, 0x00, 0x2E, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x2E, 0x2E, 0x24, 0x2E, 0x2E, 0x2E, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x2E, 0x24, 0x2E, 0x00, 0x2E, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x2E, 0x24, 0x24, 0x24, 0x24, 0x2E, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x24, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x00, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x24, 0x24, 0x24, 0x24, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 0x2E, 
  0x2E, 0x2E, 0x2E, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 
  0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x24, 0x2E, 0x42, 0x00]
print(len(land))
map=''

for i in range(1830):
    if land[i]==0x41:
        map+='A'
        print('A',end='')
    elif land[i]!=0x24 & land[i]!=0x2E & land[i]!=0x00:
        map+=chr(land[i])
        print(chr(land[i]),end='')
    elif land[i]==0x24:
        map+='1'
        print('1',end='')
    elif land[i]==0x2E:
        map+='25开学考迷宫.py'
        print(0,end='')
    else:
        map+='\n'
        print('\n')
print(map)

https://i-blog.csdnimg.cn/direct/9f28ec118591483d9fa2a0052bc57b6e.png 这里为了方便观看，我将0改为了“.(点)”，更便于分辨路径和墙

第二种类型（数字型地图）

数字型地图也就是我们最初提到的0和1组成的地图

https://i-blog.csdnimg.cn/direct/5e1da1e31eb340e29403010f73e139a1.png

处理好了地图，我们就得看看原文件的关键函数，找到各方向的输入字符

寻找关键函数

一个小tip，就是也用shift+f12去查看字符，看到中文或者类似意思的英文去定位回其地址，然后就得到了关键函数 https://i-blog.csdnimg.cn/direct/f99ede3ef3c5407a93a9daf1f5db4c21.png 比如这里双击后可以回到汇编语言的界面 https://i-blog.csdnimg.cn/direct/4dd8efb10f7c4b60a14fa1a57f6254c9.png 这里双击其函数名，就又到了思维导图似的页面 https://i-blog.csdnimg.cn/direct/bbd5ad19c0ab422eb8ccf9f0e505f34d.png

__int64 sub_140011980()
{
  char *v0; // rdi
  __int64 i; // rcx
  char v3[32]; // [rsp+0h] [rbp-20h] BYREF
  char v4; // [rsp+20h] [rbp+0h] BYREF
  int v5; // [rsp+24h] [rbp+4h]
  int v6; // [rsp+44h] [rbp+24h]
  int v7; // [rsp+64h] [rbp+44h]
  int v8; // [rsp+84h] [rbp+64h]
  unsigned int v9; // [rsp+A4h] [rbp+84h]
  unsigned int v10; // [rsp+C4h] [rbp+A4h]
  char v11[32]; // [rsp+E4h] [rbp+C4h] BYREF
  unsigned int v12; // [rsp+104h] [rbp+E4h]
  unsigned int v13; // [rsp+124h] [rbp+104h]
  char v14; // [rsp+1F4h] [rbp+1D4h]

  v0 = &v4;
  for ( i = 74i64; i; --i )
  {
    *(_DWORD *)v0 = -858993460;
    v0 += 4;
  }
  sub_14001137A(&unk_14002400E);
  v5 = 0;
  v6 = 0;
  v7 = 59;
  v8 = 29;
  v9 = 0;
  v10 = 0;
  sub_14001119A(byte_14001AD30);
  do
  {
    while ( 1 )
    {
      while ( 1 )
      {
        sub_14001119A("请输入移动指令 > ");
        sub_140011203(" %c", v11);
        v12 = v9;
        v13 = v10;
        v14 = v11[0];
        if ( v11[0] == 67 )
        {
          --v12;
          goto LABEL_14;
        }
        if ( v14 == 88 )
        {
          ++v13;
          goto LABEL_14;
        }
        if ( v14 == 89 )
          break;
        if ( v14 == 90 )
        {
          ++v12;
          goto LABEL_14;
        }
        sub_14001119A(asc_14001ADE0);
      }
      --v13;
LABEL_14:
      if ( v12 <= 0x3C && v13 < 0x1F )
        break;
      v9 = 0;
      v10 = 0;
      sub_14001119A(asc_14001AE08);
    }
    if ( aA[61 * v13 + v12] == 36 )
    {
      v9 = 0;
      v10 = 0;
      sub_14001119A(asc_14001AE38);
    }
    else
    {
      v9 = v12;
      v10 = v13;
      sub_14001119A("你移动到了 (%d, %d)\n", v12, v13);
    }
  }
  while ( v9 != 59 || v10 != 29 );
  sub_14001119A(asc_14001AE90);
  sub_140011316(v3, &unk_14001AD00);
  return 0i64;
}

ai分析得到方向 https://i-blog.csdnimg.cn/direct/5c7e48845fac480992f2bcc2ae9a5c09.png

此段代码看到下方有一个“v9 != 59 || v10 != 29”，这就说明了我们的地图大小为60*30

路径解决

深度优先搜索

什么是DFS

深度优先搜索，即DFS，一种用于遍历或搜索树或图的算法。 沿着树的深度遍历树的节点，尽可能深的搜索树的分支。当节点v的所在边都己被探寻过或者在搜寻时结点不满足条件，搜索将回溯到发现节点v的那条边的起始节点。整个进程反复进行直到所有节点都被访问为止。

一个形象的演示 https://i-blog.csdnimg.cn/direct/01b02d8e14bc447ab9a1a2fe30333c97.gif 深度优先搜索一般是通过递归来实现的，“递”的过程就是往下搜的过程对应着深度，“归”的过程就是回溯，回退上一级

基本思路（特点）
  1. 不论道路通或不通以及通向何处，都是一直朝一个方向往深处探索的，直到撞墙
  2. 遇到墙后才往回走，且只回溯到上一个节点，然后选择其他的方向行进
  3. 从出发点一直深入，遇墙回溯，直到到达终点结束

宽度优先搜索

什么是BFS

宽度优先搜索，即BFS。C++中的 （Breadth-First Search，BFS）是一种图遍历算法，它从图中的一个特定顶点开始，按照“广度”优先的顺序逐层遍历图中的节点。该算法通常用于寻找最短路径、图的连通性以及状态空间搜索等领域。

所谓广度，就是一层一层的，向下遍历，层层堵截。

详细来讲：从初始节点开始，应用算符生成第一层节点，检查目标节点是否在这些后继节点中，若没有，再用产生式规则将所有第一层的节点逐一扩展，得到第二层节点，并逐一检查第二层节点中是否包含目标节点。若没有，再用算符逐一扩展第二层所有结点……，如此依次扩展，直到发现目标结点为止 。

基本思路（特点）
  1. 一层一层的遍历，按层遍历，同层的都试完后进入下一层
  2. 所有点分为已访问的点和未访问的点，已访问的不会再次访问
  3. 一旦找到终点立刻结束，这样就能实现最短路径

对于BFS和DFS比较形象的一个比较 https://i-blog.csdnimg.cn/direct/5c9faebb420048c2bbed0a870b5df4b5.jpeg

上面的那个例子需要的是最短路径，所以采用BFS

写脚本


maze=[['0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1'],['1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1'],['1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1'],['1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0'],['1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '0', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0'],['1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', 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'1', '1', '1', '1', '1', '1', '1', '1', '0', '0']]
from collections import deque #使用 deque（双端队列）来实现广度优先搜索（BFS）。deque 提供了高效的队列操作，支持从头部和尾部快速添加和删除元素。


def bfs(maze, start, end):  #定义bfs函数
    rows, cols = len(maze), len(maze[0])  #从maze中提取地图每行每列的元素的个数
    directions = [(0, -1, 'C'), (0, 1, 'Z'), (1, 0, 'X')] #定义三个可能移动的方向的代码，C,Z,X
    queue = deque([start]) 
    visited = set([start]) #使用deque存储已访问的点
    prev = {}

    while queue:
        x, y = queue.popleft()
        if (x, y) == end:
            break
        for dx, dy, d in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < rows and 0 <= ny < cols and maze[nx][ny] == '0' and (nx, ny) not in visited:
                queue.append((nx, ny))
                visited.add((nx, ny))
                prev[(nx, ny)] = (x, y, d)

    if end not in prev:
        return None

    path = []
    x, y = end
    while (x, y) != start:
        x, y, d = prev[(x, y)]
        path.append(d)
    path.reverse()

    return path

#定义路径查找的函数

def find_shortest_path(maze):
    rows, cols = len(maze), len(maze[0])
    start = (0, 0)
    end = (rows - 1, cols - 1)
    path = bfs(maze, start, end)
    return path

path = find_shortest_path(maze)
print("".join(path) if path else "No path found")

https://i-blog.csdnimg.cn/direct/bf56056cbc1a43e5a1db051efef0f680.png 得到后通常还需要进行md5加密（结果为32位） https://i-blog.csdnimg.cn/direct/8bec43d0e4a94bbaaf297b9aab62520b.png

更新于 2025-03-14
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