（POJ2251） Dungeon Master <3维 BFS>

Dungeon Master
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!

Source Ulm Local 1997

题意：
有一个3维的迷宫，问你从S出发到E最少需要多少步？

分析：
简单的BFS题
WA了两次。。。第一次是因为m[][][]的顺序是z,x,y，我在处理的时候用的x,y,z
第二次是没有注释掉freopen()…

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;

char m[35][35][35];
char ss[35];
int L,R,C;
int d[6][3] = {0,-1,0,0,1,0,1,0,0,-1,0,0,0,0,-1,0,0,1};
int sx,sy,sz;
bool v[35][35][35];
int ans;

struct node
{
    int x,y,z;
    int step;
};

bool bfs()
{
    memset(v,0,sizeof(v));
    node now,ne;
    queue<node> q;
    while(!q.empty()) q.pop();
    now.x = sx; now.y = sy; now.z = sz; now.step = 0;
    v[sx][sy][sz]=1;
    q.push(now);
    while(!q.empty())
    {
        now = q.front(); q.pop();
        if(m[now.z][now.x][now.y] == 'E')//注意m的顺序。。
        {
            ans = now.step;
            return true;
        }
        int tx,ty,tz;
        for(int i=0;i<6;i++)
        {
            tx = now.x+d[i][0];
            ty = now.y+d[i][1];
            tz = now.z+d[i][2];
            if(tx >= 0 && tx < R && ty >= 0 && ty <C && tz >=0 && tz < L)
            {
                if(!v[tx][ty][tz] && m[tz][tx][ty]!='#')//注意m的顺序。。
                {
                    ne.x = tx;
                    ne.y = ty;
                    ne.z = tz;
                    ne.step = now.step+1;
                    v[tx][ty][tz]=1;
                    q.push(ne);
                }
            }
        }
    }
    return false;
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d%d",&L,&R,&C)!=EOF)
    {
        if(L==0 && R==0 && C==0) break;
        for(int i=0;i<L;i++)
        {
            for(int j=0;j<R;j++)
            {
                //gets(m[i][j]);
                scanf("%s",m[i][j]);
                for(int k=0;k<C;k++)
                {
                    if(m[i][j][k]=='S')
                    {
                        sx = j;
                        sy = k;
                        sz = i;
                    }
                }
            }
            gets(ss);
        }
//        for(int i=0;i<L;i++)
//        {
//            for(int j=0;j<R;j++)
//                cout<<m[i][j]<<endl;
//        }
        if( bfs()) printf("Escaped in %d minute(s).\n",ans);
        else printf("Trapped!\n");

    }
    return 0;
}