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
- 本地navicatl连接linux
首选你Linux服务器上要装配好了MySQL数据库.输入: # mysql -u root -proot mysql>GRANT ALL PRIVILEGES ON *.* TO 'root'@ ...
- iOS开发 Swift开发数独游戏(四) 游戏界面的界面与逻辑
一.游戏界面涉及到的功能点 1)数独格子的建模 (1)绘制数独格子要考虑到标记功能 所以要在每个格子内预先塞入9个标记数字,仅数独格子算下来就有9*9*9=729个格子且存在大量嵌套(这导致我在操作S ...
- 如何在AutoCAD中将卫星底图变为有坐标参考信息的
这篇博文首先没有图,主要是博主太懒了,不想再截图,我把过程说清楚也可以的.特此说明. (1)将下载好的瓦片拼接好大的地图 (2)将其导入到ArcGIS中,定义其地理坐标,如WGS84:然后将其其投影为 ...
- window linux 文件传输
window 安装:pscp.exe (放在C:\Windows\System32 目录下) Linux 安装: 1: 先更新apt-getroot@ubuntu:/home/ubuntu# sudo ...
- Qt中的非模式窗口配置;
Test7_5A::Test7_5A(QWidget *parent) : QMainWindow(parent){ ui.setupUi(this); m_searchwin = new Searc ...
- C#中out与ref区别
一.ref(参考)与out区别 1.out(只出不进) 将方法中的参数传递出去,在方法中将该参数传递出去之前需要在该方法起始赋初值:在方法外传递的该参数可以不用赋值: 简单理解就是:将一个东西抛出去之 ...
- python git log
# -*- coding: utf-8 -*- # created by vince67 Feb.2014 # nuovince@gmail.com import re import os imp ...
- 【原创】SpringBoot & SpringCloud 快速入门学习笔记(完整示例)
[原创]SpringBoot & SpringCloud 快速入门学习笔记(完整示例) 1月前在系统的学习SpringBoot和SpringCloud,同时整理了快速入门示例,方便能针对每个知 ...
- 倍福TwinCAT(贝福Beckhoff)基础教程1.2 TwinCAT安装配置
由于TC2和TC3都有可能用到,个人推荐都安装,但是注意必须是先安装的TwinCAT2,然后安装TwinCAT3,如果反了可能两个都没法用(打开TcSwitchRuntime提示Both TwinCA ...
- Win8.1应用开发之Bing Maps
这里介绍怎样进行Bing Maps的开发.首先我们须要在我们的程序中引入Bing Map的SDK.详细方法,这里推荐一个链接<win8>使用Bing地图.这样一个hello world便出 ...