POJ 2251:Dungeon Master(三维BFS)

标签：poj   bfs   

Description

Input

Output

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;
}

POJ 2251:Dungeon Master(三维BFS),布布扣,bubuko.com

POJ 2251:Dungeon Master(三维BFS)

标签：poj   bfs   

原文地址：http://blog.csdn.net/u013487051/article/details/37901353