Crashing Robots - poj 2632
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8352 | Accepted: 3613 |
Description
A robot crashes with a wall if it attempts to move outside the area of the warehouse, and two robots crash with each other if they ever try to occupy the same spot.
Input
The second line contains two integers, 1 <= N, M <= 100, denoting the numbers of robots and instructions respectively.
Then follow N lines with two integers, 1 <= Xi <= A, 1 <= Yi <= B and one letter (N, S, E or W), giving the starting position and direction of each robot, in order from 1 through N. No two robots start at the same position.
Figure 1: The starting positions of the robots in the sample warehouse
Finally there are M lines, giving the instructions in sequential order.
An instruction has the following format:
< robot #> < action> < repeat>
Where is one of
- L: turn left 90 degrees,
- R: turn right 90 degrees, or
- F: move forward one meter,
and 1 <= < repeat> <= 100 is the number of times the robot should perform this single move.
Output
- Robot i crashes into the wall, if robot i crashes into a wall. (A robot crashes into a wall if Xi = 0, Xi = A + 1, Yi = 0 or Yi = B + 1.)
- Robot i crashes into robot j, if robots i and j crash, and i is the moving robot.
- OK, if no crashing occurs.
Only the first crash is to be reported.
Sample Input
4
5 4
2 2
1 1 E
5 4 W
1 F 7
2 F 7
5 4
2 4
1 1 E
5 4 W
1 F 3
2 F 1
1 L 1
1 F 3
5 4
2 2
1 1 E
5 4 W
1 L 96
1 F 2
5 4
2 3
1 1 E
5 4 W
1 F 4
1 L 1
1 F 20
Sample Output
Robot 1 crashes into the wall
Robot 1 crashes into robot 2
OK
Robot 1 crashes into robot 2
#include <iostream>
#include<string.h>
using namespace std; int main() { int K=;
int ew,ns;
int robots_num;
int instruction;
//读取记录机器人的位置
int robots_posx[];
int robots_posy[];
int robots_to[];
//读取记录对机器人的操作
int ins_rob[];
char ins_op[];
int ins_rep[];
//记录所有机器人的位置
int pos[][];
cin>>K;
//flag 记录状态 -1 Ok;0 撞到墙;其他 撞到机器人的编号
int flag=-;
int rob_n;
for(int i=;i<K;i++){
flag=-;
rob_n=;
memset(pos,,sizeof(int)*);
cin>>ew>>ns>>robots_num>>instruction;
for(int j=;j<robots_num;j++){
char tmp;
cin>>robots_posx[j]>>robots_posy[j];
pos[robots_posy[j]-][robots_posx[j]-]=j+;
cin>>tmp;
switch(tmp){
case('N'):
robots_to[j]=;
break;
case('E'):
robots_to[j]=;
break;
case('S'):
robots_to[j]=;
break;
case('W'):
robots_to[j]=;
break;
}
}
for(int j=;j<instruction;j++){
cin>>ins_rob[j];
cin>>ins_op[j];
cin>>ins_rep[j];
}
for(int j=;j<instruction;j++){
int rob=ins_rob[j]-;
char op=ins_op[j];
if(op=='L')
robots_to[rob]=(robots_to[rob]+(-ins_rep[j]%))%;
if(op=='R')
robots_to[rob]=(robots_to[rob]+ins_rep[j])%;
if(op=='F'){
int rep=ins_rep[j];
pos[robots_posy[rob]-][robots_posx[rob]-]=;
int tow=robots_to[rob];
if(tow==){
for(int k=;k<rep;k++){
robots_posy[rob]++;
if(robots_posy[rob]>ns){
flag=;
rob_n=rob+;
break;
}else if(pos[robots_posy[rob]-][robots_posx[rob]-]!=){
flag=pos[robots_posy[rob]-][robots_posx[rob]-];
rob_n=rob+;
break;
}
}
}else if(tow==){
for(int k=;k<rep;k++){
robots_posx[rob]++;
if(robots_posx[rob]>ew){
flag=;
rob_n=rob+;
break;
}else if(pos[robots_posy[rob]-][robots_posx[rob]-]!=){
flag=pos[robots_posy[rob]-][robots_posx[rob]-];
rob_n=rob+;
break;
}
}
}else if(tow==){
for(int k=;k<rep;k++){
robots_posy[rob]--;
if(robots_posy[rob]<){
flag=;
rob_n=rob+;
break;
}else if(pos[robots_posy[rob]-][robots_posx[rob]-]!=){
flag=pos[robots_posy[rob]-][robots_posx[rob]-];
rob_n=rob+;
break;
}
}
}else if(tow==){
for(int k=;k<rep;k++){
robots_posx[rob]--;
if(robots_posx[rob]<){
flag=;
rob_n=rob+;
break;
}else if(pos[robots_posy[rob]-][robots_posx[rob]-]!=){
flag=pos[robots_posy[rob]-][robots_posx[rob]-];
rob_n=rob+;
break;
}
}
}
if(flag!=-)
break;
pos[robots_posy[rob]-][robots_posx[rob]-]=rob+;
} }
if(flag==-)
cout<<"OK"<<endl;
else if(flag==)
cout<<"Robot "<<rob_n<<" crashes into the wall"<<endl;
else
cout<<"Robot "<<rob_n<<" crashes into robot "<<flag<<endl;
}
return ;
}
Crashing Robots - poj 2632的更多相关文章
- Crashing Robots POJ 2632 简单模拟
Description In a modernized warehouse, robots are used to fetch the goods. Careful planning is neede ...
- 模拟 POJ 2632 Crashing Robots
题目地址:http://poj.org/problem?id=2632 /* 题意:几个机器人按照指示,逐个朝某个(指定)方向的直走,如果走过的路上有机器人则输出谁撞到:如果走出界了,输出谁出界 如果 ...
- Poj OpenJudge 百练 2632 Crashing Robots
1.Link: http://poj.org/problem?id=2632 http://bailian.openjudge.cn/practice/2632/ 2.Content: Crashin ...
- POJ 2632 Crashing Robots (坑爹的模拟题)
Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6599 Accepted: 2854 D ...
- poj 2632 Crashing Robots
点击打开链接 Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6655 Accepted: ...
- poj 2632 Crashing Robots(模拟)
链接:poj 2632 题意:在n*m的房间有num个机器,它们的坐标和方向已知,现给定一些指令及机器k运行的次数, L代表机器方向向左旋转90°,R代表机器方向向右旋转90°,F表示前进,每次前进一 ...
- POJ 2632:Crashing Robots
Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8424 Accepted: 3648 D ...
- Crashing Robots 分类: POJ 2015-06-29 11:44 10人阅读 评论(0) 收藏
Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8340 Accepted: 3607 D ...
- 模拟 --- Crashing Robots
Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7394 Accepted: 3242 D ...
随机推荐
- Java Socket编程详细解说
Java Socket编程 JavaSocketServerSocket乱码超时 Java Socket编程 对于Java Socket编程而言,有两个概念,一个是ServerSocket,一个是So ...
- 在linux下给grep命令添加颜色
1打开文件,添加如下一段话 vim ~/.bashrc alias grep='grep --color' 2退出保存 source ~/.bashrc ...
- CentOS 6.9安装类型选择(Basic Server/Web Server)
Desktop :基本的桌面系统,包括常用的桌面软件,如文档查看工具. Minimal Desktop:基本的桌面系统,包含的软件更少. Minimal:基本的系统,不含有任何可选的软件包. Basi ...
- PHP下载/采集远程图片到本地
/** * 下载远程图片到本地 * * @param string $url 远程文件地址 * @param string $filenNme 保存后的文件名(为空时则为随机生成的文件名,否则为原文件 ...
- powerdesigner操作
转:http://w-tingsheng.blog.163.com/blog/static/25056034201392331452977/ 使用PowerDesigner生成数据库建表SQL脚 本时 ...
- Mybaits的特点及优点
1.SQL语句和代码分离 便于统一管理和维护,不必再Java代码中调式SQL语句.但是当SQL语句出错时,控制台不会打印Log, 解决办法--引用log4j 2.用标签拼接SQL语句 用标签代替JAV ...
- 自定义ViewPager的兼容性问题及解决办法
通过它我们可以给图片增加组合动画效果,也可以写成一个图片查看器. 比如我们首次安装应用的时候,很多就会用到ViewPager给我们做一个应用简介.今天要写的也是这个--怎么用ViewPager实现动画 ...
- android SQLite(单词的添加与查询应用)
本人小白,刚接触android,为方便记忆,将平时练习的代码写下来,跟大家分享,也希望大神批评指正. 这个实例主要用到的SQLite数据库的操作,可以向数据库添加单词,查询,修改以及删除单词,描述如有 ...
- autoRelease
cocos2dx采用的是引用计数的方式来管理对象的持有和释放. 所谓引用计数就是说,每个对象都会有一个属性用来记录当前被几个地方引用了.在释放内存的时候会根据这个引用计数来确定是否要用delete操作 ...
- react-native 常用组件的用法(二)
ScrollView组件 能够调用移动平台的ScrollView(滚动视图)的组件,同时还集成了触摸锁定的“响应者”系统.注意一定要给scrollview一个高度,或者是他父级的高度. 常用方法 on ...