poj 2251 Dungeon Master 3维bfs(水水)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 21230 | Accepted: 8261 |
Description
Is an escape possible? If yes, how long will it take?
Input
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
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
- 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 ...
- 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 ...
- POJ 2251 Dungeon Master【三维BFS模板】
Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 45743 Accepted: 17256 Desc ...
- 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& ...
- 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 ...
- 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 ...
- POJ 2251 Dungeon Master(地牢大师)
p.MsoNormal { margin-bottom: 10.0000pt; font-family: Tahoma; font-size: 11.0000pt } h1 { margin-top: ...
- 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 ...
- 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
- linux 命令——47 iostat (转)
Linux系统中的 iostat 是I/O statistics(输入/输出统计)的缩写,iostat工具将对系统的磁盘操作活动进行监视.它的特点是汇报磁盘活动统计情况,同时也会 汇报出CPU使用情况 ...
- POI 怎么设置Excel整列的CellStyle啊
POI 怎么设置Excel整列的CellStyle啊,而不是循环每个Cell.因为现在是生成Excel模板,不知道客户会输入多少行. 问题补充: 指尖言 写道 好像没有这个方法,CellStyle是C ...
- World Wind Java开发之六——解析shape文件(转)
http://blog.csdn.net/giser_whu/article/details/41647117 最近一直忙于导师项目的事情了,几天没更新了,昨天和今天研究了下WWJ解析shp文件的源代 ...
- 使用OpenFileDialog组件打开对话框
实现效果: 知识运用: OpenFileDialog组件的ShowDialog方法 public DialogResult Show () //返回枚举值 DialogRrsult.OK 或 Di ...
- 2017年9月22日作业 c++算术运算符 自增 自减 逻辑运算符 位运算符 条件运算符(三元运算符)
作业1: c++算术运算符试题,分析下面程序的输出结果是什么 //第一个: int x=8999;int value=x*1000/1000; //第二个 int x=8999;int value=x ...
- Android驱动开发5-7总结
Android深度探索5-7章总结 介绍了S3C6410开发板的功能,开发板的不同主要是在烧录嵌入式系统的方式不同,以及如何在此开发板上安装Android.紧接着学到介绍到如何在多种平台,使用多种方式 ...
- AIDE
安装 yum install aide 修改配置文件 vim /etc/aide.conf (指定对哪些文件进行检测) /test/chameleon R /bin/ps R+a /usr/bin/c ...
- 致敬wusir懒孩子自有懒孩子的生存之道之二
https://www.cnblogs.com/wupeiqi/ https://www.cnblogs.com/Eva-J/ https://www.cnblogs.com/wupeiqi/p/90 ...
- 503. Next Greater Element II
https://leetcode.com/problems/next-greater-element-ii/description/ class Solution { public: vector&l ...
- Codeforces Round #459 (Div. 2):B. Radio Station
B. Radio Station time limit per test2 seconds memory limit per test256 megabytes Problem Dsecription ...