题意：
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!
以前都是二位搜索，三维bfs就有上下左右前后六个位置，其他处理没区别。
import java.util.ArrayDeque; import java.util.Comparator; import java.util.PriorityQueue; import java.util.Queue; import java.util.Scanner; /* * 0 1 1 0 * 0 0 0 0 * 0 0 0 0 * 0 1 1 0 */ public class testB { static int xstart=0;static int ystart=0;static int hstart=0; static int xend=0;static int yend=0;static int hend=0; static int x1[]= {0,1,0,0,-1,0};//上下左右 前后 static int y1[]= {1,0,0,-1,0,0}; static int z1[]= {0,0,1,0,0,-1}; public static void main(String[] args) { // TODO 自动生成的方法存根 Scanner sc=new Scanner(System.in); while(sc.hasNext()) { int l=sc.nextInt();//lever层数 h int r=sc.nextInt();//行 x int c=sc.nextInt();//列 y if(l==0&&r==0&&c==0)break; char map[][][]=new char[l][r][c]; for(int i=0;i<l;i++) { for(int j=0;j<r;j++) { map[i][j]=sc.next().toCharArray(); } // sc.next(); } for(int i=0;i<l;i++) { for(int j=0;j<r;j++) { for(int k=0;k<c;k++) { if(map[i][j][k]=='S') { xstart=j;ystart=k;hstart=i; } } // System.out.println(); } //System.out.println(" "); } boolean jud[][][]=new boolean[l][r][c]; Queue<node>q1=new ArrayDeque<node>(); q1.add(new node(xstart, ystart, hstart,0)); boolean b=false;int x,y,z=0; jud[hstart][xstart][ystart]=true; while(!q1.isEmpty()) { node point=q1.poll(); x=point.x;y=point.y; z=point.h; for(int i=0;i<6;i++) { if(x+x1[i]<r&&x+x1[i]>=0&&y+y1[i]>=0&&y+y1[i]<c&&z+z1[i]>=0&&z+z1[i]<l)//没有越界 { if(map[z+z1[i]][x+x1[i]][y+y1[i]]=='.'&&!jud[z+z1[i]][x+x1[i]][y+y1[i]]) { q1.add(new node(x+x1[i], y+y1[i], z+z1[i], point.minitu+1)); jud[z+z1[i]][x+x1[i]][y+y1[i]]=true; } else if(map[z+z1[i]][x+x1[i]][y+y1[i]]=='E') { System.out.println("Escaped in "+(point.minitu+1)+" minute(s)."); b=true; break; } } } if(b)break; } if(!b) { System.out.println("Trapped!"); } } } static Comparator<node>com=new Comparator<node>() { public int compare(node o1, node o2) { // TODO 自动生成的方法存根 return (int)(o1.minitu-o2.minitu); } }; } class node { int x;int y;int h; int minitu; public node(int x,int y,int h, int minitu) { this.x=x;this.y=y;this.h=h;this.minitu=minitu; } }
