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POJ 2251:Dungeon Master(三维BFS)

2024-07-12 10:58:38   220人阅读

Dungeon Master


Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 16178
Accepted: 6268

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!



3维的迷宫，只需要加上上下两个方向就行了，其他的就是简单的bfs。

开始就是在判断那个进队的if语句中。。应该是map[xx][yy][zz]!=‘#‘, 开始写成了map[xx][yy][zz]==‘.‘；所以害我调了一会。。


#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>

using namespace std;

const int N = 105;

char map[N][N][N];
int vist[N][N][N];

struct node
{
    int x;
    int y;
    int z;
    int num;
};
queue<node>q;

int dx[]={1,-1,0,0,0,0};
int dy[]={0,0,0,0,-1,1};
int dz[]={0,0,-1,1,0,0};
int flag;
int t, n, m;
int xx, yy, zz;

void bfs()
{
    flag = 0;
    while( !q.empty() )
    {
        node temp = q.front();
        //cout<<map[temp.x][temp.y][temp.z]<<" ";
        //printf( "%d %d %d\n", temp.x, temp.y, temp.z );
        if( map[temp.x][temp.y][temp.z]=='E' )
        {
            printf("Escaped in %d minute(s).\n", temp.num);
            flag = 1;
            //printf("1\n");
            break;
        }
        q.pop();
        for( int i=0; i<6; i++ )
        {
            xx = temp.x + dx[i];
            yy = temp.y + dy[i];
            zz = temp.z + dz[i];
            //printf( "%d %d %d\n" ,xx,yy,zz);
            if( xx>=1 && xx<=t && yy>=1 && yy<=n &&zz>=1 && zz<=m && !vist[xx][yy][zz] && map[xx][yy][zz]!='#' )
            {
                node next;
                next.x = xx; next.y = yy; next.z = zz; next.num = temp.num+1;
                //printf( "%d %d %d\n", xx, yy, zz );
                vist[xx][yy][zz] = true;
                q.push( next );
            }
        }
    }
    if( !flag )
        printf("Trapped!\n");
}

int main()
{
    while( scanf("%d%d%d", &t, &n, &m) &&t &&n &&m )
    {
        node frist;
        while( !q.empty() )  q.pop();
        memset( vist, false, sizeof( vist ) );
        for(int i=1; i<=t; i++)
            for(int j=1; j<=n; j++)
                for(int k=1; k<=m; k++)
                {
                    cin>>map[i][j][k];
                    if( map[i][j][k]=='S' )
                    {
                        frist.x = i;
                        frist.y = j;
                        frist.z = k;
                    }
                }
       frist.num=0;
       vist[frist.x][frist.y][frist.z] = true;
       q.push( frist );
       bfs();
    }

    return 0;
}











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