POJ - 2251 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!

题意：一个立体的图形，你要找到他的出口用最短的时间，S是入口，E是出口，注意别写错。。还有一些细节我在下面的代码注释中写出来。

#include <queue>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
char s[35][35][35];
int dx, dy, dz, ex, ey, ez;
int visited[35][35][35];
int sx[] = {1, -1, 0, 0, 0, 0};
int sy[] = {0, 0, 1, -1, 0, 0};
int sz[] = {0, 0, 0, 0, 1, -1};
int l, r, c;
struct position
{
    int x, y, z;
    int step;
};
bool inside(int x, int y, int z)
{
    return x >= 0 && y >= 0 && z >= 0 && x < l && y < r && z < c;
}
int bfs()
{
    queue<position> q;
    position p, p1, p2;
    p.x = dx;
    p.y = dy;
    p.z = dz;
    visited[dx][dy][dz] = 1;
    p.step = 0;
    q.push(p);
    memset(visited, 0, sizeof(visited));
    while(!q.empty())
    {
        p1 = q.front();
        q.pop();
        int x = p1.x;
        int y = p1.y;
        int z = p1.z;
        if(visited[x][y][z] == 1) //这句话必须写上，否贼就会超时，这点我改了1个多小时才发现因为这里的一些问题超时
            continue;
        visited[x][y][z] = 1;
        if(x == ex && y == ey && z == ez)
        {
            return p1.step;
        }
        for(int i = 0; i < 6; i++)
        {
            int xx = p1.x + sx[i];
            int yy = p1.y + sy[i];
            int zz = p1.z + sz[i];
            if(inside(xx, yy ,zz) && visited[xx][yy][zz] == 0&&s[xx][yy][zz] != '#')  // 判断是否越界，是否符合条件
            {
                p2.x = xx;
                p2.y = yy;
                p2.z = zz;
                p2.step = p1.step + 1;
                q.push(p2);
            }
        }

    }
    return -1;
}
int main()
{
    while(scanf("%d%d%d", &l, &r, &c) != EOF)
    {
        memset(visited, -1, sizeof(visited));
        if(l == 0 && r == 0 && c== 0)
            break;
        for(int i = 0; i < l ;i++)
        {
            for(int j = 0;j < r; j++)
                scanf("%s", s[i][j]);
        }
        for(int i = 0; i < l; i++)
        {
            for(int j = 0; j < r; j++)
            {
                for(int k = 0; k < c; k++)
                {
                    if(s[i][j][k] == 'S')  // 记录入口的横纵竖坐标
                      dx = i, dy = j, dz = k;
                      if(s[i][j][k] == 'E') // 记录出口的横竖纵坐标
                        ex = i, ey = j, ez = k;
                }
            }
        }
        int ans =  bfs();
        if(ans != -1)
            cout << "Escaped in " << ans << " minute(s)."<<endl;
        else
            cout << "Trapped!" << endl;

    }
    return 0;
}