QAQ……queue Q，這操作困擾了我很久

首先，分享一下我悲劇的經歷。聽說DFS容易出各種各樣莫（ren)名(pin)其(bu)妙(hao)的錯，然後BFS各種好，然後就去看了BFS（這就是我悲劇的開端）。先是在網上看別人的代碼，然後看到一個：

queue<node> Q

這樣的寫法，很迷，還特地把代碼打下來，研究了兩個星期。

然後……

一無所獲，之後就去諮詢了學長，然後學長告訴我，這個大概是之前定義了結構體。這個時候才發現我看的代碼是核心代碼(╯‵□′)╯︵┻━┻  (╯‵□′)╯︵┻━┻。

然後分享一下題目吧

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 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

然後是別的大佬的代碼（不知道是哪位大佬的，反正在 http://blog.csdn.net/libin56842/article/details/23702395 能看到）

#include <iostream>  
#include <stdio.h>  
#include <string.h>  
#include <queue>  
#include <algorithm>  
using namespace std;  
  
char map[35][35][35];  
int vis[35][35][35];  
int k,n,m,sx,sy,sz,ex,ey,ez;  
int to[6][3] = {{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};  
  
struct node  
{  
    int x,y,z,step;  
};  
  
int check(int x,int y,int z)  
{  
    if(x<0 || y<0 || z<0 || x>=k || y>=n || z>=m)  
        return 1;  
    else if(map[x][y][z] == '#')  
        return 1;  
    else if(vis[x][y][z])  
        return 1;  
    return 0;  
}  
  
int bfs()   ///廣搜 
{  
    int i;  
    node a,next;   
    queue<node> Q;  ///node 結構體 
    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(i = 0; i<6; i++)  
        {  
            next = a;  
            next.x = a.x+to[i][0];  
            next.y = a.y+to[i][1];  
            next.z = a.z+to[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()  
{  
    int i,j,r;  
    while(scanf("%d%d%d",&k,&n,&m),n+m+k)  
    {  
        for(i = 0; i<k; i++)  
        {  
            for(j = 0; j<n; j++)  
            {  
                scanf("%s",map[i][j]);  
                for(r = 0; r<m; r++)  
                {  
                    if(map[i][j][r] == 'S')  
                    {  
                        sx = i,sy = j,sz = r;  
                    }  
                    else if(map[i][j][r] == 'E')  
                    {  
                        ex = i,ey = j,ez = r;  
                    }  
                }  
            }  
        }  
        memset(vis,0,sizeof(vis));  
        int ans;  
        ans = bfs();  
        if(ans)  
            printf("Escaped in %d minute(s).\n",ans);  
        else  
            printf("Trapped!\n");  
    }  
  
    return 0;  
}