POJ 2251 Dungeon Master ——————BFS

最新推荐文章于 2021-07-30 15:28:23 发布

陶鸿杰

最新推荐文章于 2021-07-30 15:28:23 发布

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分类专栏： BFS+DFS

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本文介绍了一个3D迷宫逃脱问题的经典解决方法，通过使用广度优先搜索算法（BFS），实现寻找从起点到终点的最短路径。文章提供了一段C++代码示例，详细展示了如何遍历迷宫并记录最短路径。

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Dungeon Master

Language:Default

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 48826		Accepted: 18436

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!

Source

Ulm Local 1997

—– 虽然搜索的是三维的物体，但是是方向只有6个，也很简单。。。没啥说的，一切尽在代码中 ——

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 33;
char str[MAXN][MAXN][MAXN];
bool vis[MAXN][MAXN][MAXN];
int sx,sy,sz,L,R,C;
int d[6][3]={1,0,0, -1,0,0, 0,1,0, 0,-1,0, 0,0,1, 0,0,-1};
struct node{
    int x,y,z,step;
    node(){};
    node(int _x,int _y,int _z,int _step)
    {
        x=_x; y=_y; z=_z; step=_step;
    }
    bool operator < (const node &b)const
    {
       return step>b.step;
    }
};

int bfs(int x,int y,int z)
{
    memset(vis,0,sizeof(vis));
    vis[z][x][y]=1;
    int ans=-1;
    priority_queue<node>    que;
    que.push(node(x,y,z,0));
    node e1,e2;
    while(que.size())
    {
        e1=que.top();que.pop();
        if(str[e1.z][e1.x][e1.y]=='E')
        {
            ans=e1.step;
            break;
        }
        for(int i=0;i<6;i++)
        {
            e2.x=e1.x+d[i][0];
            e2.y=e1.y+d[i][1];
            e2.z=e1.z+d[i][2];
            e2.step=e1.step+1;
            if(!vis[e2.z][e2.x][e2.y] && str[e2.z][e2.x][e2.y]!='#' && 0 <= e2.z && e2.z < L && 0 <= e2.x && e2.x < R && 0 <= e2.y && e2.y < C)
            {
                vis[e2.z][e2.x][e2.y]=1;
                que.push(e2);
            }
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%d %d %d",&L,&R,&C)&&(L|R|C))
    {
        for(int k=0;k<L;k++)
            for(int i=0;i<R;i++)
                scanf("%s",str[k][i]);
        int z=0;
        for(int k=0;k<L;k++)
        {
            if(z)   break;
            for(int i=0;i<R;i++)
            {
                if(z)   break;
                for(int j=0;j<C;j++)
                    if(str[k][i][j]=='S')
                    {
                        sx=i;
                        sy=j;
                        sz=k;
                        z=1;
                        break;
                    }
            }
        }
        int ans=bfs(sx,sy,sz);
        if(ans==-1)    puts("Trapped!");
        else           printf("Escaped in %d minute(s).\n",ans);
    }
    return 0;
}