hdu 1240 3维迷宫 求起点到终点的步数 (BFS)

题意，给出一个N，这是这个三空间的大小，然后给出所有面的状况O为空地,X为墙，再给出起始点的三维坐标和终点的坐标，输出到达的步数

比较坑 z是x，x是y，y是z，
Sample Input
START 1
O
0 0 0
0 0 0
END
START 3
XXX
XXX
XXX
OOO
OOO
OOO
XXX
XXX
XXX
0 0 1
2 2 1
END
START 5
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
XXXXX
XXXXX
XXXXX
XXXXX
XXXXX
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
0 0 0
4 4 4
END

Sample Output
1 0
3 4
NO ROUTE

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <queue>
 5 using namespace std;
 6 
 7 char map[20][20][20];
 8 int v[20][20][20];
 9 int n;
10 int sx,sy,sz;
11 int ex,ey,ez;
12 int dx[] = {1,-1,0,0,0,0};
13 int dy[] = {0,0,1,-1,0,0};
14 int dz[] = {0,0,0,0,1,-1};
15 int ans ;
16 
17 struct node
18 {
19     int x , y ,z ,step ;
20 };
21 
22 int bfs()
23 {
24     node now , t ;
25     int fx , fy , fz ;
26     queue<node> q ;
27     now.x = sx ;
28     now.y = sy ;
29     now.z = sz ;
30     now.step = 0 ;
31     memset(v,0,sizeof(v)) ;
32     q.push(now) ;
33     v[sz][sx][sy] = 1 ;
34     while(!q.empty())
35     {
36         now = q.front() ;
37         q.pop() ;
38         if (now.x == ex && now.y == ey && now.z == ez)
39         {
40             ans  = now.step ;
41             return  1 ;
42         }
43         for (int i = 0 ; i < 6 ; i++)
44        {
45         fx = now.x + dx[i] ;
46         fy = now.y + dy[i] ;
47         fz = now.z + dz[i] ;
48         if (fx<0 || fy<0 || fz<0 || fx>=n || fy>=n || fz>=n || v[fz][fx][fy] == 1 || map[fz][fx][fy] == 'X')
49             continue ;
50 
51         t.x = fx ;
52         t.y = fy ;
53         t.z = fz ;
54         t.step = now.step + 1 ;
55         v[fz][fx][fy] = 1 ;
56         q.push(t) ;
57         }
58     }
59 
60     return 0 ;
61 }
62 
63 int main ()
64 {
65     //freopen("in.txt","r",stdin) ;
66     char s[10] ;
67     while (scanf("%s %d" , s , &n) !=EOF)
68     {
69         int i , j , k ;
70         for (i = 0 ; i < n ; i ++)
71             for (j = 0 ; j < n ; j ++)
72                   {
73                       scanf("%s" , map[i][j]) ;
74                   }
75         scanf("%d %d %d" , &sx,&sy,&sz) ;
76         scanf("%d %d %d" , &ex,&ey,&ez) ;
77         scanf("%s" , s) ;
78 
79         ans = 0 ;
80         if (bfs())
81             printf("%d %d\n" , n , ans) ;
82         else
83             printf("NO ROUTE\n") ;
84     }
85 
86     return 0 ;
87 }
View Code

 

posted @ 2015-05-19 22:18  __Meng  阅读(308)  评论(0)  编辑  收藏  举报