Dungeon Master

Dungeon Master
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 20995 Accepted: 8150

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
三维的搜索,开始少算了时间复杂度,用dfs超时,果断的改成bfs

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#define WW freopen("output.txt","w",stdout)

const int MAX = 10000;

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

struct point
{
    int num;
    node a;
};

int L,R,C;

int Dir[][3]= {{0,1,0},{0,-1,0},{1,0,0},{-1,0,0},{0,0,1},{0,0,-1}};

char Map[35][35][35];

node s,e;

bool vis[35][35][35];

int Max;

bool Judge(int x,int y,int z)
{
    if(x<R&&x>=0&&y<C&&y>=0&&z<L&&z>=0&&vis[z][x][y]==false&&Map[z][x][y]!='#')
    {
        return true;
    }
    return false;
}

void bfs()
{
    point b,c;
    queue<point>Q;
    memset(vis,false,sizeof(vis));
    b.a.x=s.x;
    b.a.y=s.y;
    b.a.z=s.z;
    b.num=0;
    vis[s.z][s.x][s.y]=true;
    Q.push(b);
    while(!Q.empty())
    {
        c=Q.front();
        Q.pop();
        if(c.a.x==e.x&&c.a.y==e.y&&c.a.z==e.z)
        {
            Max=c.num;
            return ;
        }
        for(int i=0;i<6;i++)
        {
            b.a.x=c.a.x+Dir[i][0];
            b.a.y=c.a.y+Dir[i][1];
            b.a.z=c.a.z+Dir[i][2];
            b.num=c.num+1;
            if(Judge(b.a.x,b.a.y,b.a.z))
            {
                vis[b.a.z][b.a.x][b.a.y]=true;
                Q.push(b);
            }

        }
    }

}
int main()
{
    while(~scanf("%d %d %d",&L,&R,&C))
    {
        if(L==0&&R==0&&C==0)
        {
            break ;
        }
        for(int i=0; i<L; i++)
        {
            for(int j=0; j<R; j++)
            {
                scanf("%s",Map[i][j]);
                for(int k=0; k<C; k++)
                {
                    if(Map[i][j][k]=='S')
                    {
                        s.x=j;
                        s.y=k;
                        s.z=i;
                    }
                    if(Map[i][j][k]=='E')
                    {
                        e.x=j;
                        e.y=k;
                        e.z=i;
                    }
                }
            }
        }
        Max = INF;
        bfs();
        if(Max<INF)
        {
            printf("Escaped in %d minute(s).\n",Max);
        }
        else
            printf("Trapped!\n");
    }
    return 0;
}