深度优先搜索（DFS）和广度优先搜索（BFS）求解迷宫问题

用下面这个简单的迷宫图作为例子：

OXXXXXXX
OOOOOXXX
XOXXOOOX
XOXXOXXO
XOXXXXXX
XOXXOOOX
XOOOOXOO
XXXXXXXO

O为通路，X为障碍物。

深度优先搜索就像是一条路走到黑，走到黑，黑了再回来。有种递归的感觉。

深度优先搜索（DFS）

 #include<iostream>
 using namespace std;

 char a1[] = {'O','X','X','X','X','X','X','X','\0'};
 char a2[] = {'O','O','O','O','O','X','X','X','\0'};
 char a3[] = {'X','O','X','X','O','O','O','X','\0'};
 char a4[] = {'X','O','X','X','O','X','X','O','\0'};
 char a5[] = {'X','O','X','X','X','X','X','X','\0'};
 char a6[] = {'X','O','X','X','O','O','O','X','\0'};
 char a7[] = {'X','O','O','O','O','X','O','O','\0'};
 char a8[] = {'X','X','X','X','X','X','X','O','\0'};
 char *p[] = {a1, a2, a3, a4, a5, a6, a7, a8};

 int offset[][] = {{-, }, {, }, {, -}, {, }};　　//偏移量为一位大小，顺序为：上（x-1）、下（x+1）、左(y-1)、右(y+1)

 void DFS(int x, int y)
 {
     if(x ==  && y == )　　　//找到出口
     {
         p[x][y] = '*';
         cout << "One path of the maze:" << endl;
         for(int i = ; i < ; i++)
             cout << p[i] << endl;
         cout << endl;
     }
     for(int i = ; i < ; i++)　　//寻找可行的路径
     {
         int nx = x + offset[i][];　　//试探可走路径，下一位置为当前位置加上偏移量
         int ny = y + offset[i][];
         if(nx>= && nx<= && ny>= && ny<= && p[nx][ny]!='X' && p[nx][ny]!='*')　　//找到可行路径
         {
             p[x][y] = '*';　　//画出路径
             DFS(nx, ny);　　//继续搜索
             p[x][y] = 'O';　　//走不下去了回退
         }
     }
 }

 int main()
 {
     cout << "The size of the maze： row 8， col 8" << endl;
     cout << "The map: " << endl;
     for(int i = ; i < ; i++)
         cout << p[i] << endl;

     DFS(, );
     return ;
 }

广度优先搜索则是遍历与当前位置相邻的所有可行点，就像是病毒，传播速度很快。一传十，十传百的感觉。
求解时需要与队列相结合。

广度优先搜索（BFS）

 #include<iostream>
 #include<cstring>
 using namespace std;

 char a1[] = {'O','X','X','X','X','X','X','X','\0'};
 char a2[] = {'O','O','O','O','O','X','X','X','\0'};
 char a3[] = {'X','O','X','X','O','O','O','X','\0'};
 char a4[] = {'X','O','X','X','O','X','X','O','\0'};
 char a5[] = {'X','O','X','X','X','X','X','X','\0'};
 char a6[] = {'X','O','X','X','O','O','O','X','\0'};
 char a7[] = {'X','O','O','O','O','X','O','O','\0'};
 char a8[] = {'X','X','X','X','X','X','X','O','\0'};
 char *p[] = {a1, a2, a3, a4, a5, a6, a7, a8};    

 int offset[][] = {{-, }, {, }, {, -}, {, }};
 int vis[][]; //用来标记是否访问过当前位置
 int cnt = ;

 struct Position{
     int x, y;    //当前位置
     int pre;    //前驱点
 }path[*], myqueue[*];

 void BFS()
 {
     memset(vis, , sizeof(vis));
     int front = , rear = ;

     myqueue[rear].x = ;    //入口入队
     myqueue[rear].y = ;
     myqueue[rear].pre = -;
     rear++;

     Position* tmp;
     while(front < rear)
     {
         tmp = &myqueue[front];    //当前位置出队 

         if(tmp->x ==  && tmp->y == )    //到达出口
         {
             while(tmp->pre != -)    //回溯寻找路径
             {
                 path[cnt].x = tmp->x;　　//记录可行路径的位置
                 path[cnt].y =tmp->y;
                 cnt++;
                 tmp = &myqueue[tmp->pre];
             }

             for(int i = ; i < cnt; i++)　　//在地图中可视化路径
             {
                 p[][] = '*';
                 p[path[i].x][path[i].y] = '*';
             }
             cout << "One path of the maze:" << endl;
             for(int i = ; i < ; i++)
                 cout << p[i] << endl;
             cout << endl;
             break;
         }

         for(int i = ; i < ; i++)    //遍历相邻点
         {
             int nx = tmp->x + offset[i][];　　//试探路径
             int ny = tmp->y + offset[i][];
             if(nx>= && nx<= && ny>= && ny<= && p[nx][ny]!='X' && vis[nx][ny]!=)    //满足条件入队
             {
                 vis[nx][ny] = ;
                 myqueue[rear].x = nx;
                 myqueue[rear].y = ny;
                 myqueue[rear].pre = front;
                 rear++;
             }
         }
         front++;
     }
 }

 int main()
 {
     cout << "The size of the maze： row 8， col 8" << endl;
     cout << "The map: " << endl;
     for(int i = ; i < ; i++)
         cout << p[i] << endl;

     BFS();
     return ;
 }

这是一个只有一条通路的迷宫，具体要根据需求进行修改以满足。