Dungeon Master（poj 2251)

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.
Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. If it is not possible to escape, print the line

Trapped!

Sample Input

3 4 5

S....

.###.

.##..

###.#



#####

#####

##.##

##...



#####

#####

#.###

####E



1 3 3

S##

#E#

###



0 0 0

Sample Output

Escaped in 11 minute(s).

Trapped!


题意：3维迷宫！“.”是路；“#”是墙，“E"进“S”出
这题的题目略有小坑；
注意几个地方：1.数组开到105以上；2.队列清空；

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cstring>

 4 #include<iostream>

 5 #include<queue>

 6 using namespace std;

 7 int s[133][133][133];

 8 int vis[133][133][133];

 9 int r[6]= {-1,1,0,0,0,0};

10 int c[6]= {0,0,-1,1,0,0};

11 int h[6]= {0,0,0,0,1,-1};

12 int ei,ej,ek;

13 struct node

14 {

15     int a,b,c;

16     int step;

17 };

18 node p,q;

19 void getit(int I,int J,int K)//输入并处理

20 {

21     char ch;

22     int i,j,k;

23     for(k=1; k<K; k++)

24     {

25         for(i=1; i<I; i++)

26         {

27             for(j=1; j<J; j++)

28             {

29                 scanf("%c",&ch);

30                 if(ch=='.')

31                     s[i][j][k]=-1;

32                 else if(ch=='E')

33                 {

34                     ei=i;

35                     ej=j;

36                     ek=k;

37                 }

38                 else if(ch=='S')

39                     s[i][j][k]=1;

40             }

41             getchar();

42         }

43         getchar();

44     }

45 }

46 void bfs(int ai,int aj, int ak,int step)//简单BFS

47 {

48     queue<node>que;

49     vis[ai][aj][ak]=1;

50     int t;

51     p.a=ai;

52     p.b=aj;

53     p.c=ak;

54     p.step=step;

55     que.push(p);

56     while(!que.empty())

57     {

58         p=que.front();

59         que.pop();

60         for(t=0; t<6; t++)

61         {

62             q.a=p.a+r[t];

63             q.b=p.b+c[t];

64             q.c=p.c+h[t];

65             q.step=p.step+1;

66             if(s[q.a][q.b][q.c]&&!vis[q.a][q.b][q.c])

67             {

68                 if(s[q.a][q.b][q.c]==1)

69                 {

70                     printf("Escaped in %d minute(s).\n",q.step);

71                     return ;

72                 }

73                 que.push(q);

74                 vis[q.a][q.b][q.c]=1;

75             }

76         }

77     }

78     printf("Trapped!\n");

79 }

80 int main()

81 {

82 

83     int I,J,K;

84     while(scanf("%d %d %d",&K,&I,&J)&&(I||J||K))

85     {

86         getchar();

87         memset(s,0,sizeof(s));//置空

88         memset(vis,0,sizeof(vis));

89         getit(I+1,J+1,K+1);

90         bfs(ei,ej,ek,0);

91     }

92     return 0;

93 }

View Code