走迷宫BFS算法 

 1 #include<iostream>
 2 #include<queue>
 3 #include<stack>
 4 #include<iomanip>
 5 using namespace std;
 6 
 7 int map[40][40];
 8 int done[40][40];
 9 
10 int dx[4] = { 0,0,1,-1 };
11 int dy[4] = { 1,-1,0,0 };
12 
13 struct Node
14 {
15     int x, y, step;
16 };
17 
18 int main() {
19     int n,ex,ey;
20     cin >> n>>ex>>ey;
21 
22     for (int i = 1; i <= n; i++)
23         for (int j = 1; j <= n; j++)
24             done[i][j] = -1;
25 
26     for (int i = 1; i <= n; i++)
27         for (int j = 1; j <= n; j++)
28             cin >> map[i][j];
29     
30     queue<Node> q;
31     q.push(Node({1,1,0}));
32     done[1][1] = 0;
33     while (!q.empty())
34     {
35         //cout << q.front().x << ":" << q.front().y << ":" << q.front().step << endl;
36         for (int i = 0; i < 4; i++)
37         {
38             int tx = q.front().x + dx[i];
39             int ty = q.front().y + dy[i];
40             
41             if (tx>=1&&ty>=1&&tx<=n&&ty<=n&&done[tx][ty]==-1&&!map[tx][ty])
42             {
43                 q.push(Node({ tx,ty,q.front().step+1 }));
44                 done[tx][ty] = q.front().step + 1;
45             }
46         }
47         q.pop();
48     }
49 
50     cout << endl;
51 
52     for (int i = 1; i <= n; i++) {
53         for (int j = 1; j <= n; j++) {
54             if (done[i][j] != -1) cout << left <<setw(2) << done[i][j];
55             else cout << left << setw(2) << "□";
56         }
57         cout << endl;
58     }
59     
60     for (int i = 0; i <= n+1; i++)
61         for (int j = 0; j <= n + 1; j++)
62             if (i == 0 || i == n + 1 || j == 0 || j == n + 1) done[i][j] = -10;
63     int ttx = ex, tty = ey;
64     int step = done[tty][ttx];
65     stack<char> result;
66     while (step)
67     {
68         step--;
69         if (done[tty][ttx-1] == step) {
70             //cout << "L"; 
71             ttx--; result.push('R');
72         }else
73         if (done[tty][ttx+1] == step) {
74             //cout << "R"; 
75             ttx++; result.push('L');
76         }else
77         if (done[tty-1][ttx] == step) {
78             //cout << "U"; 
79             tty--; result.push('D');
80         }else
81         if (done[tty+1][ttx] == step) {
82             //cout << "D"; 
83             tty++; result.push('U');
84         }
85     }
86     while (!result.empty())
87     {
88         cout << result.top();
89         result.pop();
90     }
91 
92     return 0;
93 }

运用技术：BFS算法/栈结构和队列结构

posted @ 2020-08-31 22:46  时光潜流  阅读(313)  评论(0)  编辑  收藏  举报