3D dungeon（BFS)

描述 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?

输入

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.

输出

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!

样例输入

3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 0 0 0

样例输出

Escaped in 11 minute(s). Trapped!

该题大意：给一个多层地图，#为墙，从S出发走到E处，若能输出最短时间（每次只能上下左右，上一层，下一层移动，一次1分钟），若不能则输出Trapped!。

思路：可以理解成可向6个方向走的bfs，用3位数组存储地图。（还是需要勤加练习啊，实在是菜到家了）。

代码如下：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
char a[100][100][100];
int vis[100][100][1000];//访问数组，判断是否访问过
int k,n,m,i,j,h;
int f[6][3]={{0,1,0},{0,0,1},{0,-1,0},{0,0,-1},{-1,0,0},{1,0,0}};//6个可走的方向
struct node
{int x,y,z,t;
} s,em;
int bfs(int x1,int y1,int z1,int t)
{queue<node>q;s.x=x1;s.y=y1;s.z=z1;vis[x1][y1][z1]=1;s.t=t;q.push(s);while(!q.empty()){s=q.front();q.pop();for(i=0;i<6;i++){em.x=s.x+f[i][0];em.y=s.y+f[i][1];em.z=s.z+f[i][2];if(em.x>=0&&em.x<k&&em.y>=0&&em.y<n&&em.z>=0&&em.z<m&&vis[em.x][em.y][em.z]==0&&a[em.x][em.y][em.z]!='#')//判别下一步的条件{em.t=s.t+1;if(a[em.x][em.y][em.z]=='E'){return em.t;}vis[em.x][em.y][em.z]=1;//标记已走过改点q.push(em);}}}return -1;
}
int main()
{while(scanf("%d%d%d",&k,&n,&m)!=EOF){if(k==0&&n==0&&m==0){break;}memset(vis,0,sizeof(vis));int x1,y1,z1;for(i=0;i<k;i++){for(j=0;j<n;j++){for(h=0;h<m;h++){cin>>a[i][j][h];//注意：如果用scanf输入需要用getchar吸收回车if(a[i][j][h]=='S'){x1=i;y1=j;z1=h;}}}}int fag;fag=bfs(x1,y1,z1,0);if(fag==-1){printf("Trapped!\n");}else{printf("Escaped in %d minute(s).\n",fag);}}return 0;
}

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