POJ 2251 Dungeon Master(三维6方向BFS)

B - Dungeon Master

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2251

Appoint description:  System Crawler  (2015-03-28)

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!

题目大意：

　　给你一个３维的迷宫，维数是L*R*C,告诉了你起始的地点和终止的地点，每次只能够向6个方向来走动，问是否能够走出去，如果能的话，求出这个

最小的时间。

解题思路：

　　直接BFS上最短时间问题，len[rear] = len[front]+1;每次进队前都要更新路径就好了。

代码：

# include<cstdio>
# include<iostream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>

using namespace std;

const double PI=4.0*atan(1.0);

typedef long long LL;
typedef unsigned long long ULL;

# define inf 999999999
# define MAX 33

char grid[MAX][MAX][MAX];
int book[MAX][MAX][MAX];
int len[MAX*MAX*MAX];
int l,r,c;
int ans;
int st_x,st_y,st_z;
int ed_x,ed_y,ed_z;

int next[6][3] = {{1,0,0},{0,1,0},{-1,0,0},{0,-1,0},{0,0,1},{0,0,-1}};

struct node
{
    int x;
    int y;
    int z;
}que[MAX*MAX*MAX];

void input()
{
        for ( int i = 0;i < l;i++ )
        {
            for ( int j = 0;j < r;j++ )
            {
                for ( int k = 0;k < c;k++ )
                {
                    char ch;
                    ch = getchar();
                    grid[i][j][k] = ch;
                    if ( ch=='S' )
                    {
                        st_x = i;
                        st_y = j;
                        st_z = k;
                    }
                }
                getchar();
            }
            getchar();
        }
}

int bfs()
{
    int front,rear;
    memset(book,0,sizeof(book));
    memset(len,0,sizeof(len));
    que[0].x = st_x;
    que[0].y = st_y;
    que[0].z = st_z;
    rear = front = 0;
    while ( front <= rear )
    {
        for ( int i = 0;i < 6;i++ )
        {
            int n_x = que[front].x+next[i][0];
            int n_y = que[front].y+next[i][1];
            int n_z = que[front].z+next[i][2];

            if ( book[n_x][n_y][n_z]==0&&(grid[n_x][n_y][n_z]=='.'||grid[n_x][n_y][n_z]=='E')&&n_x>=0&&n_x<l&&n_y>=0&&n_y<r&&n_z>=0&&n_z<c )
            {
                book[n_x][n_y][n_z] = 1;
                que[++rear].x = n_x;
                que[rear].y = n_y;
                que[rear].z = n_z;
                len[rear] = len[front]+1;
                if ( grid[n_x][n_y][n_z] == 'E' )
                {
                    return len[rear];
                }
            }

        }
        front++;
    }
    return 0;

}

int main(void)
{
    while ( cin>>l>>r>>c )
    {
        if ( l==0&&r==0&&c==0 )
            break;
            getchar();
        input();
        ans = bfs();
        if ( ans )
            printf("Escaped in %d minute(s).\n",ans);
        else
            printf("Trapped!\n");
    }

    return 0;
}