Dungeon Master(POJ--2251

最新推荐文章于 2022-10-14 19:40:10 发布

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最新推荐文章于 2022-10-14 19:40:10 发布

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分类专栏： --------------简单搜索------------ BFS ------------POJ-杂七杂八----------

本文链接：https://blog.youkuaiyun.com/zhang_di233/article/details/47731567

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BFS

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本文介绍了一种解决三维迷宫问题的广度优先搜索（BFS）算法，旨在找到从入口到出口的最短路径并计算所需时间。通过解析输入的三维地图描述，算法能够判断是否存在逃脱路线，并计算出逃离所需的最短时间。若无法逃脱，则输出'Trapped!'。

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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!

题意：这是一个三维的地图，“S”为入口，“E”为出口，“#”表示不能走，“.”表示能走，问是否能从入口走到出口，如果能则输出所需最短时间（走一步需花一分钟），如果不能则输出“Trapped!”.

思路：直接用BFS进行一个三维的广搜即可。

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define MAX 0x3f3f3f3f
using namespace std;
int h,n,m,sz,sx,sy,ez,ex,ey,ans;
char mmap[40][40][40];    //记录三维地图
bool vis[40][40][40];          //标记是否走过
int dis[40][40][40];             //记录到该点所花的时间
int dz[]= {1,-1,0,0,0,0};
int dx[]= {0,0,1,-1,0,0};
int dy[]= {0,0,0,0,1,-1};
typedef struct node
{
    int z,x,y;
} Node;
int BFS()
{
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    queue<Node>q;               //用结构体入队
    Node nd;
    nd.z=sz;
    nd.x=sx;
    nd.y=sy;
    q.push(nd);
    vis[sz][sx][sy]=true;
    while(!q.empty())
    {
        int xx=q.front().x;
        int yy=q.front().y;
        int zz=q.front().z;
        q.pop();
        if(xx==ex&&yy==ey&&zz==ez)
        {
            ans=dis[zz][xx][yy];
            return 1;
        }
        for(int i=0; i<6; i++)
        {
            int xxx=xx+dx[i];
            int yyy=yy+dy[i];
            int zzz=zz+dz[i];
            if(xxx>=0&&xxx<n&&yyy>=0&&yyy<m&&zzz>=0&&zzz<h&&!vis[zzz][xxx][yyy]&&mmap[zzz][xxx][yyy]!='#')
            {
                dis[zzz][xxx][yyy]=dis[zz][xx][yy]+1;
                vis[zzz][xxx][yyy]=true;
                nd.z=zzz;
                nd.x=xxx;
                nd.y=yyy;
                q.push(nd);
            }
        }
    }
    return 0;
}
int main()
{
    //freopen("lalala.text","r",stdin);
    while(~scanf("%d %d %d",&h,&n,&m))
    {
        if(h==0&&n==0&&m==0)
            break;
        for(int i=0; i<h; i++)
            for(int j=0; j<n; j++)
            {
                scanf("%s",mmap[i][j]);
                getchar();
            }
        for(int i=0; i<h; i++)
            for(int j=0; j<n; j++)
                for(int k=0; k<m; k++)
                {
                    if(mmap[i][j][k]=='S')
                    {
                        sz=i;
                        sx=j;
                        sy=k;
                    }
                    if(mmap[i][j][k]=='E')
                    {
                        ez=i;
                        ex=j;
                        ey=k;
                    }
                }
        if(BFS())
            printf("Escaped in %d minute(s).\n",ans);
        else
            printf("Trapped!\n");
    }
    return 0;
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