poj-2251-Dungeon Master(bfs)

最新推荐文章于 2022-10-14 19:40:10 发布

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本文介绍了一种解决三维迷宫最短路径问题的方法，利用广度优先搜索算法找到从起点到终点的最快路径。

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23841		Accepted: 9270

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!

题意：给出三维空间的地牢,要求求出由字符'S'到字符'E'的最短路径移动方向可以是上，下，左，右，前，后，六个方向每移动一次就耗费一分钟，要求输出最快的走出时间。（上下楼梯时，xy坐标不变）

思路：就是把方向增加到6个，其他没什么区别，水过。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <queue>
#define INF 0x3f3f3f3f
#define CSH(a, b) memset(a, (b), sizeof(a))
using namespace std;
char s[31][31][31];
int dir[6][3]={-1,0,0, 1,0,0, 0,-1,0, 0,1,0, 0,0,1, 0,0,-1};
int did[2]={-1,1};
int t,m,n;
int flag;
struct node
{
    int x,y,z,ti;
}fr;
void intt()
{
    for(int i=0;i<t;i++)
    {
        for(int j=0;j<n;j++)
        {
            getchar ();
            for(int k=0;k<m;k++)
            {
                scanf("%c",&s[i][j][k]);
                if(s[i][j][k]=='S')
                {
                    fr.x=j;
                    fr.y=k;
                    fr.z=i;
                    fr.ti=0;
                    s[i][j][k]='#';
                }
                
            }
        }
        getchar();
    }
    
}
void bfs()
{
    queue<node> q;
    while(!q.empty())
    {
        q.pop();
    }
    q.push(fr);
    while(!q.empty())
    {
        node ne;
        ne=q.front();
        q.pop();
        for(int i=0;i<6;i++)
        {
            node no;
            no.x=ne.x+dir[i][0];
            no.y=ne.y+dir[i][1];
            no.z=ne.z+dir[i][2];
            no.ti=ne.ti;
            if(no.x>=0&&no.x<n&&no.y<m&&no.y>=0&&no.z<t&&no.z>=0&&s[no.z][no.x][no.y]!='#')//判断边界
            {
                no.ti++;
                if(s[no.z][no.x][no.y]=='E')
                {
                    flag=no.ti;
                    break;
                }
                s[no.z][no.x][no.y]='#';
                //操作..
                q.push(no);
            }
            if(flag)
                break;
        }
    }
    
}
int main()
{
    while(~scanf("%d%d%d",&t,&n,&m)&&n+m+t)
    {
        flag=0;
        intt();
        bfs();
        if(flag)
            printf("Escaped in %d minute(s).\n",flag);
        else
            printf("Trapped!\n");
    }
}