POJ 3009 Curling 2.0 dfs

题目： http://poj.org/problem?id=3009

很简单的一个题，调试了好久。。

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 int n, m, maze[30][30], ans;
 5 int start_x, start_y, end_x, end_y;
 6 int dirc[4][2] = {{0,1}, {1,0}, {0,-1}, {-1,0}};
 7 
 8 void dfs(int x, int y, int step, int dir)
 9 {
10     if(step > 10 || maze[x][y] == -1)return;
11     if(x == end_x && y == end_y)
12     {
13         if(ans == -1 || step < ans)
14             ans = step;
15         return;
16     }
17     int nx = x + dirc[dir][0];
18     int ny = y + dirc[dir][1];
19     if(maze[nx][ny] != 1)
20     {
21         dfs(nx, ny, step, dir);
22         return;
23     }
24     else if(maze[nx][ny] == 1)
25     {
26         maze[nx][ny] = 0;
27         for(int d = 0; d < 4; d++)
28         {
29             int xx = x + dirc[d][0];
30             int yy = y + dirc[d][1];
31             if(maze[xx][yy] != 1)
32                 dfs(xx, yy, step+1, d);
33         }
34         maze[nx][ny] = 1;
35     }
36 }
37 
38 int main()
39 {
40     while(scanf("%d %d", &m, &n) != EOF)
41     {
42         if(n == 0 && m == 0)break;
43         memset(maze, -1, sizeof(maze));
44         for(int i = 1; i <= n; i++)
45         {
46             for(int j = 1; j <= m; j++)
47             {
48                 scanf("%d", &maze[i][j]);
49                 if(maze[i][j] == 2)
50                 {
51                     start_x = i;
52                     start_y = j;
53                 }
54                 if(maze[i][j] == 3)
55                 {
56                     end_x = i;
57                     end_y = j;
58                 }
59             }
60         }
61         ans = -1;
62         for(int d = 0; d < 4; d++)
63         {
64             int nx = start_x + dirc[d][0];
65             int ny = start_y + dirc[d][1];
66             if(maze[nx][ny] != 1)
67                 dfs(nx, ny, 1, d);
68         }
69         printf("%d\n", ans);
70     }
71     return 0;
72 }
View Code

 

posted @ 2013-08-19 10:20  Anti-Magic  阅读(172)  评论(0)  编辑  收藏  举报