C - Dungeon Master
C - Dungeon Master
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Is an escape possible? If yes, how long will it take?
Input
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! //用dfs做的,就是在试探的时候多试探一次上下位置的时候就行。
#include <iostream>
#include <queue>
#include <string.h>
using namespace std; struct point
{
int x,y,h;
int step;
};
point star,end; char map [][][];
bool way [][][];
int h,x,y; void read_map()//读地图
{
for (int i=;i<=h;i++)
{
for (int j=;j<=x;j++)
{
for (int k=;k<=y;k++)
{
cin>>map[i][j][k];
if (map[i][j][k]=='S')
{
star.h=i;
star.x=j;
star.y=k;
star.step=;
}
if (map[i][j][k]=='E')
{
end.h=i;
end.x=j;
end.y=k; }
}
}
}
} int check(point t)//检查是否能走
{ if ( t.x>= && t.x<=x && t.y>= && t.y<=y && t.h>= && t.h<=h && way[t.h][t.x][t.y]== )
{
if ( map[t.h][t.x][t.y]!='#')
{
return ;
}
}
return ;
} int bfs()
{
point now,next;
int min=-; queue<point> Q;
Q.push(star);
way[star.h][star.x][star.y]=; while (!Q.empty())
{
now=Q.front();
Q.pop();
if (now.x==end.x&&now.y==end.y&&now.h==end.h)
{
min=now.step;
break;
} next.x=now.x+;
next.y=now.y;
next.h=now.h;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;} next.x=now.x;
next.y=now.y-;
next.h=now.h;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;} next.x=now.x-;
next.y=now.y;
next.h=now.h;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;} next.x=now.x;
next.y=now.y+;
next.h=now.h;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;} next.x=now.x;
next.y=now.y;
next.h=now.h+;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;} next.x=now.x;
next.y=now.y;
next.h=now.h-;
next.step=now.step+;
if (check(next)){ Q.push(next); way[next.h][next.x][next.y]=;}
}
return min; } int main()
{
int all;
while (cin>>h>>x>>y)
{
memset(way,,sizeof(way));
if (h==&&x==&&y==) break;
read_map();
all=bfs();
if (all==-)
cout<<"Trapped!"<<endl;
else
cout<<"Escaped in "<<all<<" minute(s)."<<endl;
}
return ;
}
C - Dungeon Master的更多相关文章
- POJ 2251 Dungeon Master(3D迷宫 bfs)
传送门 Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 28416 Accepted: 11 ...
- poj 2251 Dungeon Master
http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submis ...
- Dungeon Master 分类: 搜索 POJ 2015-08-09 14:25 4人阅读 评论(0) 收藏
Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20995 Accepted: 8150 Descr ...
- 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 ...
- UVa532 Dungeon Master 三维迷宫
学习点: scanf可以自动过滤空行 搜索时要先判断是否越界(L R C),再判断其他条件是否满足 bfs搜索时可以在入口处(push时)判断是否达到目标,也可以在出口处(pop时) #i ...
- Dungeon Master poj 2251 dfs
Language: Default Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16855 ...
- POJ 2251 Dungeon Master(地牢大师)
p.MsoNormal { margin-bottom: 10.0000pt; font-family: Tahoma; font-size: 11.0000pt } h1 { margin-top: ...
- BFS POJ2251 Dungeon Master
B - Dungeon Master Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u ...
- POJ 2251 Dungeon Master (非三维bfs)
Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 55224 Accepted: 20493 ...
- 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 ...
随机推荐
- Digg工程师讲述Digg背后的技术
虽然最近业绩有所下滑,也出现了一些技术故障,但Digg作为首屈一指的社会化新闻网站,其背后的技术还是值得一探,Digg工程师 Dave Beckett 在今年4月份写一篇名为<How Digg ...
- Cocos2d-x 3.0中 物理碰撞检測中onContactBegin回调函数不响应问题
好吧,事实上这篇也是暂时冒出来的,近期朋友要做个物理游戏,曾经做物理还是用box2d,呃.确实要花些功夫才干搞懂当中的精髓,可是听讲这套引擎又一次封装了一次.要easy非常多,所以就简单尝试了一下,感 ...
- 火车票订票API 用PHP完成火车票订票流程
本教程用来演示聚合数据-火车票订票接口的使用流程. 配置好PHP环境,PHP版本最好大于5.5 去聚合数据-火车票订票接口申请key:http://www.juhe.cn/docs/api/id/17 ...
- Hadoop之词频统计小实验
声明: 1)本文由我原创撰写,转载时请注明出处,侵权必究. 2)本小实验工作环境为Ubuntu操作系统,hadoop1-2-1,jdk1.8.0. 3)统计词频工作在单节点的伪分布上,至于真正实 ...
- 由于删除DBF文件报错 —— ORA-01033: ORACLE initialization or shutdown in progress
由于移动或删除DBF文件报错:ORA-01033: ORACLE initialization or shutdown in progress 原因:一般该类故障通常是由于移动文件而影响了数据库日 ...
- prototype 用法
prototype使得js面向对象使用了prototype之后,使用它里面的属性或者函数 需要new出一个对象才可以使用.否则不使用prototype,直接向对象注入 function Person( ...
- MySQL学习总结(四)数据的基本操作以及MySQL运算符和常用函数
数据库是存储数据库对象的仓库,数据库的基本对象是表,表用来存储数据.关于数据的操作也就是我们常说的CRUD,C指的是CREATE(插入数据记录).R指的是READ(查询数据记录).U指的是UPDATE ...
- MVC Controller return 格式之JsonResult、ContentResult、RedirectResult……
//语法 public class JsonResult : ActionResult public class ContentResult : ActionResult public class ...
- MySQL 使用 比较函数 INTERVAL() 函数 实现数据按区间分组
首先看一下它的定义: INTERVAL(N,N1,N2,N3,..........) INTERVAL()函数进行比较列表(N1,N2,N3等等)中的N值.该函数如果N<N1返回0,如果N< ...
- ELF解析(part one)
the contents class elf { //date structure Elf32_Ehdr ehdr; Elf32_Shdr shdr; Elf32_Phdr phdr; // void ...