POJ 2251 Dungeon Master bfs

最新推荐文章于 2019-08-02 19:19:16 发布

蒟蒻的ACMer

最新推荐文章于 2019-08-02 19:19:16 发布

阅读量1k

点赞数

CC 4.0 BY-SA版权

分类专栏：大二训练 BFS 文章标签： c++ bfs poj

本文链接：https://blog.youkuaiyun.com/zp___waj/article/details/51125319

大二训练同时被 2 个专栏收录

28 篇文章

订阅专栏

BFS

8 篇文章

订阅专栏

本文深入探讨了深度学习技术在音视频处理领域的应用，包括AI音视频处理、图像处理AR特效、音视频直播流媒体等方向，详细介绍了各种应用场景及实现方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24602		Accepted: 9525

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

3d bfs//原来不知道哪里错了盯着代码debug半天没结果重新手敲一遍就a了- -

ACcode：

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#define maxn 33
using namespace std;
char mapp[maxn][maxn][maxn];
bool vis[maxn][maxn][maxn];
int to[][3]={1,0,0,-1,0,0,0,1,0,0,-1,0,0,0,1,0,0,-1};
int n,m,l;
struct N{
    int x,y,z,t;
};
N my,no,ne;
bool ju(N a){
    if(a.x<0||a.x>=n||a.y<0||a.y>=m||a.z<0||a.z>=l||mapp[a.x][a.y][a.z]=='#'||vis[a.x][a.y][a.z])return 0;
    return 1;
}
void bfs(){
    memset(vis,false,sizeof(vis));
    queue<N>q;
    q.push(my);
    vis[my.x][my.y][my.z]=1;
    while(!q.empty()){
        no=q.front();q.pop();
        for(int i=0;i<6;++i){
            ne=no;
            ne.x+=to[i][0];
            ne.y+=to[i][1];
            ne.z+=to[i][2];
            ne.t++;
            if(ju(ne)){
                if(mapp[ne.x][ne.y][ne.z]=='E'){
                     printf("Escaped in %d minute(s).\n",ne.t);
                     return;
                }
                q.push(ne);
                vis[ne.x][ne.y][ne.z]=1;
            }
        }
    }
    printf("Trapped!\n");
}
int main(){
    while(~scanf("%d%d%d",&n,&m,&l)&&(n+m+l)){
        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
                for(int z=0;z<l;++z){
                    cin>>mapp[i][j][z];
                    if(mapp[i][j][z]=='S'){
                        my.x=i;
                        my.y=j;
                        my.z=z;
                        my.t=0;
                    }
            }
        bfs();
    }
}