迷宫问题bfs， A Knight's Journey(dfs)

迷宫问题（bfs）

  1 #include <iostream>
  2 #include <queue>
  3 #include <stack>
  4 #include <cstring>
  5 
  6 using namespace std;
  7 
  8 /*广度优先搜索*/
  9 /*将每个未访问过的邻接点进队列，然后出队列，知道到达终点*/
 10 
 11 typedef class
 12 {
 13 public:
 14     int x;
 15     int y;
 16 }coordinate;
 17 
 18 int maze[5][5];            //迷宫
 19 int road[4][2] = { { 0, -1 }, { 1, 0 }, { -1, 0 }, { 0, 1 } };
 20 coordinate pre[20][20];            //记录当前坐标的前一个坐标
 21 
 22 int visited[5][5] = { 0 };
 23 
 24 /*利用一个栈，倒序输出pre中存入的坐标*/
 25 void print()
 26 {
 27     stack<coordinate> S;
 28     coordinate rhs = { 4, 4 };
 29 
 30     while (1)
 31     {
 32         S.push(rhs);
 33         if (rhs.x == 0 && rhs.y == 0)
 34             break;
 35         rhs = pre[rhs.x][rhs.y];
 36     }
 37 
 38     while (!S.empty())
 39     {
 40         rhs = S.top();
 41         S.pop();
 42         cout << "(" << rhs.x << ", " << rhs.y << ")" << endl;
 43     }
 44 }
 45 
 46 void bfs(int x, int y)
 47 {
 48     queue<coordinate> Q;            //用来帮助广度优先搜索
 49     coordinate position;
 50 
 51     position.x = x, position.y = y;
 52 
 53     Q.push(position);
 54 
 55     while (!Q.empty())
 56     {
 57         position = Q.front();
 58         Q.pop();
 59         
 60         visited[position.x][position.y] = 1;
 61         coordinate position2;
 62 
 63         if (position.x == 4 && position.y == 4)        //如果找到终点，停止搜索
 64         {
 65             print();
 66             return;
 67         }
 68 
 69         for (int i = 0; i < 4; i++)
 70         {
 71 
 72             position2.x = position.x + road[i][0];
 73             position2.y = position.y + road[i][1];
 74 
 75             if (position2.x >= 0 && position2.x <= 4 && position2.y >= 0 && position2.y <= 4 && maze[position2.x][position2.y] == 0 && !visited[position2.x][position2.y])        //如果这个邻接点不是墙且未访问过，则进队列
 76             {
 77                 Q.push(position2);
 78                 pre[position2.x][position2.y] = position;        //记录当前坐标的前一个坐标位置
 79                 visited[position2.x][position2.y] = 1;
 80             }
 81 
 82         }
 83     }
 84 }
 85 
 86 int main()
 87 {
 88     memset(maze, 0, sizeof(pre));
 89     for (int i = 0; i < 5; i++)
 90     for (int j = 0; j < 5; j++)
 91     {
 92         cin >> maze[i][j];
 93     }
 94 
 95     //memset(pre, 0, sizeof(pre));        //现将pre初始化
 96 
 97     bfs(0, 0);
 98 
 99 
100     return 0;
101 }

A Knight's Journey

 1 #include <cstring>
 2 #include<iostream>
 3 using namespace std;
 4 
 5 char route[27][2];            //记录行驶的路线
 6 int board[9][9];        //棋盘
 7 
 8 int total = 0;
 9 int length;
10 /*dfs深度优先搜索*/
11 
12 int go[8][2] = { { -1, -2 }, { 1, -2 }, { -2, -1 }, { 2, -1 }, { -2, 1 }, { 2, 1 }, { -1, 2 }, { 1, 2 } };//字典序方向
13 
14 /*一次跳跃*/
15 void knight(int p, int q,int len ,int row, int col)            //len为行走的距离，用来判断是否遍历整个棋盘, row和col为当前坐标
16 {
17     board[row][col] = 1;                
18     route[len][0] = col + 'A' - 1, route[len][1] =  row + '0';
19     length = len;
20     /*让马分别尝试8个方向的跳跃，知道跳不动为止*/
21     
22     int x, y;
23 
24     for (int i = 0; i < 8; i++)
25     {
26 
27         x = row + go[i][0];
28         y = col + go[i][1];
29 
30         if (x >0 && x <= p && y > 0 && y <= q && board[x][y] == 0)
31             knight(p, q, len + 1, x, y);
32     }
33 
34     /*if (row - 1 >= 1 && col - 2 > 0 && board[row - 1][col - 2] == 0)
35         knight(p, q, len + 1, row - 1, col - 2);
36 
37     if (row + 1 <= p && col - 2 > 0 && board[row + 1][col - 2] == 0)
38         knight(p, q, len + 1, row + 1, col - 2);
39 
40     if (row - 2 >= 1 && col - 1 > 0 && board[row - 2][col - 1] == 0)
41         knight(p, q, len + 1, row - 2, col - 1);
42     
43     if (row + 2 <= p && col - 1 > 0 && board[row + 2][col - 1] == 0)
44         knight(p, q, len + 1, row + 2, col - 1);
45     
46     if (row - 2 >= 1 && col + 1 <= q && board[row - 2][col + 1] == 0)
47         knight(p, q, len + 1, row - 2, col + 1);
48     
49     if (row + 2 <= p && col + 1 <= q && board[row + 2][col + 1] == 0)
50         knight(p, q, len + 1, row + 2, col + 1);
51 
52     if (row - 1 >= 1 && col + 2 <= q && board[row - 1][col + 2] == 0)
53         knight(p, q, len + 1, row - 1, col + 2);
54 
55     if (row + 1 <= p && col + 2 <= q && board[row + 1][col + 2] == 0)
56         knight(p, q, len + 1, row + 1, col + 2);*/
57 
58 }
59 
60 int main()
61 {
62     int N, p, q;        //行数， 每次的宽和高
63     cin >> N;
64     
65     while (N--)
66     {
67         cin >> p >> q;
68         knight(p, q, 0, 1, 1);            //以行走了0个单元
69         cout << "Scenario #" << ++total << ":" << endl;
70         if (length == p * q - 1)
71         {
72             
73             for (int i = 0; i <= length; i++)
74                 cout << route[i][0] << route[i][1];
75             cout << endl << endl;
76         }
77         else
78             cout << "impossible" << endl << endl;
79         memset(board, 0, sizeof(board));        //清空棋盘的足迹
80         memset(route, 0, sizeof(route));        //清空路线
81     }
82 
83     return 0;
84 }

标签：int,Knight,dfs,bfs,len,&&,position,col,row
来源： https://blog.51cto.com/u_14201949/2832078