Dungeon Master bfs

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!

3維迷宮問題。每次可以向上，向下，向左，向右，向前，向後走。從起始點走到終點，求最小步數。我們可以用bfs，每次將六種情況壓入隊列（只要此位置能走而且沒有走過）。沒有什麼很特殊的情況，直接暴力搜索就行。數組我們可以多開一點，把邊界以外的都設爲岩石（不能走），這樣因爲走不出去我們就可以不用判斷邊界的情況了。

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int a[34][34][34];                //用來表示迷宮。其中-1表示岩石，0表示可以走（輸入字符的時候裝換一下就行了）
bool b[34][34][34];               //用來判斷是否走過此位置
int n,m,S;
struct f {
	int x,y,z,s;              //s表示步數
};
f start,end;                      //起點和終點
void bfs() {
	queue<f> q;
	q.push(start);
	f p,c;
	memset(b,0,sizeof(b));
	while(!q.empty()) {
		p=q.front() ;
		q.pop();
		if(p.x==end.x&&p.y==end.y&&p.z==end.z) {        //到達終點，更新終點的步數  
			end.s=p.s;
			return ;
		}
		if(a[p.x+1][p.y][p.z]==0&&b[p.x+1][p.y][p.z]==0) {
			c.x=p.x+1;
			c.y=p.y;
			c.z=p.z;
			c.s=p.s+1;
			q.push(c);
			b[p.x+1][p.y][p.z]=1;
		}
		if(a[p.x-1][p.y][p.z]==0&&b[p.x-1][p.y][p.z]==0) {
			c.x=p.x-1;
			c.y=p.y;
			c.z=p.z;
			c.s=p.s+1;
			q.push(c);
			b[p.x-1][p.y][p.z]=1;
		}
		if(a[p.x][p.y+1][p.z]==0&&b[p.x][p.y+1][p.z]==0) {
			c.x=p.x;
			c.y=p.y+1;
			c.z=p.z;
			c.s=p.s+1;
			q.push(c);
			b[p.x][p.y+1][p.z]=1;
		}
		if(a[p.x][p.y-1][p.z]==0&&b[p.x][p.y-1][p.z]==0) {
			c.x=p.x;
			c.y=p.y-1;
			c.z=p.z;
			c.s=p.s+1;
			q.push(c);
			b[p.x][p.y-1][p.z]=1;
		}
		if(a[p.x][p.y][p.z+1]==0&&b[p.x][p.y][p.z+1]==0) {
			c.x=p.x;
			c.y=p.y;
			c.z=p.z+1;
			c.s=p.s+1;
			q.push(c);
			b[p.x][p.y][p.z+1]=1;
		}
		if(a[p.x][p.y][p.z-1]==0&&b[p.x][p.y][p.z-1]==0) {
			c.x=p.x;
			c.y=p.y;
			c.z=p.z-1;
			c.s=p.s+1;
			q.push(c);
			b[p.x][p.y][p.z-1]=1;
		}

	}

}
int main() {
	int L,R,C;
	while(cin>>L>>R>>C) {
		if(L==0&&R==0&&C==0)
			break;
		char t;
		memset(a,-1,sizeof(a));                //先全部設爲岩石，輸入.時我們在變爲可以走
      		for(int i=1; i<=L; i++) {             //從1開始，邊界0爲岩石，這樣我們就不用判斷邊界了
			for(int j=1; j<=R; j++) {
				for(int k=1; k<=C; k++) {
					cin>>t;
					if(t=='S') {
						start.x=i;
						start.y=j;
						start.z=k;
						start.s=0;
					} else if(t=='.')
						a[i][j][k]=0;
					else if(t=='E') {
						end.x=i;
						end.y=j;
						end.z=k;
						a[i][j][k]=0;
					}
				}
			}
		}
		end.s=0;
		bfs();
		if(end.s!=0)
		cout<<"Escaped in "<<end.s<<" minute(s)."<<endl;
		else
		cout<<"Trapped!"<<endl;
	}
	return 0;
}