Dungeon Master
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21230   Accepted: 8261

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

[Submit]   [Go Back]   [Status]   [Discuss]

#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
bool vis[][][];
char str[][][];
int step;
int nex[][]={,,,,,-,,,,-,,,,,,,-,};
struct node{
int x,y,z;
int dis;
};
node u,v;
int x,y,z;
int bfs(){
// printf("--->%d %d %d\n",u.x,u.y,u.z);
queue<node>q;
q.push(u);
vis[u.z][u.x][u.y]=true;
while(!q.empty()){
u=q.front();
q.pop();
if(str[u.z][u.x][u.y]=='E')
return u.dis;
for(int i=;i<;i++){
v.x=u.x+nex[i][];
v.y=u.y+nex[i][];
v.z=u.z+nex[i][];
if(str[v.z][v.x][v.y]!='#'&&!vis[v.z][v.x][v.y]&&v.x>=&&v.x<x&&v.y>=&&v.y<y
&&v.z>=&&v.z<z){
vis[v.z][v.x][v.y]=true;
v.dis=u.dis+;
q.push(v);
}
} }
return ;
} int main(){ while(scanf("%d%d%d",&z,&x,&y)!=EOF){
if(x==&&y==&&z==)
break;
memset(str,,sizeof(str));
memset(vis,false,sizeof(vis));
for(int k=;k<z;k++){
for(int i=;i<x;i++){
scanf("%s",str[k][i]);
for(int j=;j<y;j++){
if(str[k][i][j]=='S')
u.x=i,u.y=j,u.z=k,u.dis=;
}
}
}
step=;
step=bfs();
if(step>)
printf("Escaped in %d minute(s).\n",step);
else
printf("Trapped!\n");
}
return ;
}

poj 2251 Dungeon Master 3维bfs(水水)的更多相关文章

  1. POJ.2251 Dungeon Master (三维BFS)

    POJ.2251 Dungeon Master (三维BFS) 题意分析 你被困在一个3D地牢中且继续寻找最短路径逃生.地牢由立方体单位构成,立方体中不定会充满岩石.向上下前后左右移动一个单位需要一分 ...

  2. POJ - 2251 Dungeon Master 多维多方向BFS

    Dungeon Master You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is ...

  3. POJ 2251 Dungeon Master【三维BFS模板】

    Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 45743 Accepted: 17256 Desc ...

  4. POJ 2251 Dungeon Master(三维空间bfs)

    题意:三维空间求最短路,可前后左右上下移动. 分析:开三维数组即可. #include<cstdio> #include<cstring> #include<queue& ...

  5. POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路)

    POJ 2251 题目大意: 给出一三维空间的地牢,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.不同L层 ...

  6. BFS POJ 2251 Dungeon Master

    题目传送门 /* BFS:这题很有意思,像是地下城,图是立体的,可以从上张图到下一张图的对应位置,那么也就是三维搜索,多了z坐标轴 */ #include <cstdio> #includ ...

  7. POJ 2251 Dungeon Master(地牢大师)

    p.MsoNormal { margin-bottom: 10.0000pt; font-family: Tahoma; font-size: 11.0000pt } h1 { margin-top: ...

  8. POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索)

    POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索) Description You ar ...

  9. POJ 2251 Dungeon Master (三维BFS)

    题目链接:http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total S ...

随机推荐

  1. JSON.parse()与JSON.stringify()

    JSON.parse() 将字符串转成JSON 举个例子 var str = '{"name":"cn","age":"2&quo ...

  2. JS.match方法 正则表达式

    match() 方法可在字符串内检索指定的值,或找到一个或多个正则表达式的匹配. 该方法类似 indexOf() 和 lastIndexOf(),但是它返回指定的值,而不是字符串的位置. <sc ...

  3. POJ 2392 Space Elevator(多重背包)

    显然塔的总高度不会超过最大的a[i],而a[i]之前的可以到达的高度 是由a值更小的块组成,所以按照a从小到大的顺序去转移. 然后就是多重背包判断存在性了,几乎和coin那题一样. 数据没coin丧病 ...

  4. WARNING you have Transparen Huge Pages..

    redis启动警告: WARNING you have Transparent Huge Pages (THP) support enabled in your kernel. This will c ...

  5. python Scraping

    http://docs.python-guide.org/en/latest/scenarios/scrape/

  6. 函数定义和调用 -------JavaScript

    本文摘要:http://www.liaoxuefeng.com/ 定义函数 在JavaScript中,定义函数的方式如下: function abs(x) { if (x >= 0) { ret ...

  7. oc数组遍历

    #import <Foundation/Foundation.h> //数组遍历(枚举)对集合中的元素依此不重复的进行遍历 int main(int argc, const char * ...

  8. C++ 容器与继承

    如果容器类型定义为基类类型,那么虽然可以把派生类装进容器中,但是不能通过容器访问派生类自己的public成员,派生类将会倍切掉,只保留派生类的基类部分: 如果把容器定义为派生类类型,那么不能把基类类型 ...

  9. GIT 团队协作快速入门使用

    GIT使用: 1.本地新建一个文件夹 git init 2.克隆远程仓库 git clone git@xxxxx.git 3.本地创建一个dev分支 (前提是服务器端已经创建好有 DEV 分支) gi ...

  10. pycharm永久激活记录

    由于上一年安装的pycharm激活时是用的激活码,有期限的,一直到今年5月4日过期,这两天顺便把版本也更新到最新,一直用的free版,到今天提醒我free快到期了,所以才狠下心来去找解决方案,目前已经 ...