Dungeon Master （简单广搜）三维地下城

                                   Dungeon Master

 

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 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

ac代码：

#include<cstdio> #include<iostream> #include<queue>#include<cstring>#include<algorithm>using namespace std;char maze[35][35][35];int vis[35][35][35];int move[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1}} ;int l,n,m,sx,sy,sz,ex,ey,ez;struct node{int x,y,z,step;};int check(int p,int q,int r){if(p<0 || p>=l || q<0 || q>=n || r<0 || r>=m)return 0;if(maze[p][q][r]=='#')return 0;if(vis[p][q][r]==1)return 0;return 1;}int bfs(){node a,next;queue<node> q;a.x=sx;a.y=sy;a.z=sz;a.step=0;vis[sx][sy][sz]=1;q.push(a);while(!q.empty()){ a=q.front();q.pop();if(a.x==ex && a.y==ey && a.z==ez){            return a.step;}/*for(dx=-1;dx<=1;dx++)            //之前用的这个形式来找下一个点，突然发现题目 for(dy=-1;dy<=1;dy++) //里说不能斜着走。。。。要好好读题 for(dz=-1;dz<=1;dz++) */for(int i = 0; i<6; i++)          {              next = a;                    //每次尝试都是在先前a的基础上进行的，因此每次循环都要             next.x = a.x+move[i][0];   //重新将next重新赋值             next.y = a.y+move[i][1];              next.z = a.z+move[i][2];              if(!check(next.x,next.y,next.z))                  continue;              vis[next.x][next.y][next.z] = 1;              next.step = a.step+1;              q.push(next);          }  } return 0;}int main(){while(scanf("%d%d%d",&l,&n,&m)!=EOF){getchar();if((l+n+m)==0) break;memset(vis,0,sizeof(vis));//保证每次输入数据时，所有的点都是未读取过 for(int i=0;i<l;i++){for(int j=0;j<n;j++){for(int k=0;k<m;k++){cin>>maze[i][j][k];if(maze[i][j][k]=='S'){sx=i;sy=j;sz=k;}if(maze[i][j][k]=='E'){ex=i;ey=j;ez=k;}}}}//printf("%d",bfs()) ;int ans;ans=bfs();                        //if(bfs()!=0)printf("Escaped in %d minute(s).\n",bfs()); 不能写成这样，这样会使用两次bfs();     if(ans!=0)printf("Escaped in %d minute(s).\n",ans);    elseprintf("Trapped!\n");}return 0;}