POJ 2251 Dungeon Master

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 12655 Accepted: 4912

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 易做图 up the dungeon. 

R and C are the number of rows and columns 易做图 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

考察点： bfs搜索

[cpp] view plaincopy

#include <stdio.h>  

#include <string.h>  

#include <math.h>  

int a[50][50][50],status[50][50][50],sum[50][50][50];  

int stax,stay,staz,endy,endx,endz,t,n,m;  

int vex[]={0,0,0,0,1,-1};  

int vey[]={0,0,-1,1,0,0};  

int vez[]={-1,1,0,0,0,0};  

char s1[100];  

struct num  

{  

    int x,y,z;  

}queue[1000000];  

int main()  

{  

    int bfs();  

    int i,j,s,k,x;  

    char c;  

    while(scanf("%d %d %d%*c",&t,&n,&m)!=EOF)  

    {  

        if(!n&&!m&&!t)  

        {  

            break;  

        }  

        for(i=0;i<=t-1;i++)  

        {  

            for(j=0;j<=n-1;j++)  

            {  

                for(x=0;x<=m-1;x++)  

                {  

                    scanf("%c",&c);  

                    if(c=='#')  

                    {  

                        a[i][j][x]=0;  

                    }else if(c=='.')  

                    {  

                        a[i][j][x]=1;  

                    }else if(c=='S')  

                    {  

                        a[i][j][x]=1;  

                        staz=i; stax=j; stay=x;  

                    }else if(c=='E')  

                    {  

                        a[i][j][x]=1;  

                        endz=i; endx=j; endy=x;  

                    }  

                }  

                getchar();  

            }  

            gets(s1);  

        }  

        k=bfs();  

        if(k!=-1)  

        {  

            printf("Escaped in %d minute(s).\n",k);  

        }else  

        {  

            printf("Trapped!\n");  

        }  

    }  

    return 0;  

}  

int bfs()  

{  

    int i,j,base,top,x,y,z,k,xend,yend,zend;  

    base=top=0;  

    queue[top].x=stax; queue[top].y=stay; queue[top++].z=staz;  

    memset(status,0,sizeof(status));  

    status[stax][stay][staz]=1;  

    memset(sum,0,sizeof(sum));  

    k=0;  

    while(base<top)  

    {  

        x=queue[base].x; y=queue[base].y; z=queue[base++].z;  

        if(x==endx&&y==endy&&z==endz)  

        {  

            k=1;  

            break;  

        }  

        for(i=0;i<=5;i++)  

        {  www.zzzyk.comwww.zzzyk.com

            xend=x+vex[i]; yend=y+vey[i]; zend=z+vez[i];  

            if(xend>=0&&xend<=n-1&¥d>=0&¥d<=m-1&&zend>=0&&zend<=t-1&&!status[xend][

补充：软件开发 , C++ ,