poj2632 Crashing Robots
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9859 | Accepted: 4209 |
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
Source
题目大意如下:给定几个机器人在一个房间里的坐标和面对的方向(东南西北),再给定几个移动(转几次方向或沿原方向前进其次),判断这些机器人是否会撞到房间墙壁或撞到互相或平安无事。
题解:模拟即可。把方向数组设成按顺时针或逆时针的顺序可缩减代码量。转方向时,注意mod4来减少要转圈数。
15775948 | ksq2013 | 2632 | Accepted | 732K | 0MS | G++ | 1985B | 2016-07-21 18:05:33 |
代码稍长:
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;//横坐标x代表WE方向,纵坐标y代表NS方向;
//const int mx[]={0,+1,-1,0,0};mov[1]->E,沿x轴向东/右移动+1;mov[2]->W,向西/左移动-1;mov[3]->N,沿x轴不移动;mov[4]->S,沿x轴不移动.
//const int my[]={0,0,0,+1,-1};mov[1]->E,沿y轴不移动;mov[2]->W,沿y轴不移动;mov[3]->N,沿y轴向北/上移动+1;mov[4]->S,沿y轴向南/下移动-1.
//顺序:东南西北;从0开始;
const int mx[]={+1,0,-1,0};
const int my[]={0,-1,0,+1};
int n,m,p,q,mp[101][101];
struct robot{
int x,y,z;
}rt[101];
struct req{
int k,rep;
char pos;
}rq[101];
void Init()
{
scanf("%d%d%d%d",&n,&m,&p,&q);
for(int i=1;i<=p;i++){
char z;
scanf("%d%d",&rt[i].x,&rt[i].y);
getchar();
z=getchar();
mp[rt[i].x][rt[i].y]=i;
switch(z){
case 'E':rt[i].z=0;break;
case 'S':rt[i].z=1;break;
case 'W':rt[i].z=2;break;
case 'N':rt[i].z=3;break;
}
}
for(int i=1;i<=q;i++){
scanf("%d",&rq[i].k);
getchar();
rq[i].pos=getchar();
scanf("%d",&rq[i].rep);
}
}
bool judge(int k)
{
if(rt[k].x<1||rt[k].x>n||rt[k].y<1||rt[k].y>m){
printf("Robot %d crashes into the wall\n",k);
return false;
}
if(mp[rt[k].x][rt[k].y]){
printf("Robot %d crashes into robot %d\n",k,mp[rt[k].x][rt[k].y]);
return false;
}
mp[rt[k].x][rt[k].y]=k;//修改移动后位置数据;
return true;
}
bool solve()
{
for(int i=1;i<=q;i++){
int k=rq[i].k;
switch(rq[i].pos){
case 'F':
for(int j=1;j<=rq[i].rep;j++){
mp[rt[k].x][rt[k].y]=0;//清空原位置数据;
rt[k].x+=mx[rt[k].z];
rt[k].y+=my[rt[k].z];
if(!judge(k))return false;
}
break;
case 'L':rt[k].z=(rt[k].z-rq[i].rep%4+4)%4;break;
case 'R':rt[k].z=(rt[k].z+rq[i].rep%4)%4;break;
}
}
return true;
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
memset(mp,0,sizeof(mp));
Init();
if(solve())puts("OK");
}
return 0;
}
poj2632 Crashing Robots的更多相关文章
- POJ2632——Crashing Robots
Crashing Robots DescriptionIn a modernized warehouse, robots are used to fetch the goods. Careful pl ...
- POJ2632 Crashing Robots 解题报告
Description In a modernized warehouse, robots are used to fetch the goods. Careful planning is neede ...
- POJ2632 Crashing Robots(模拟)
题目链接. 分析: 虽说是简单的模拟,却调试了很长时间. 调试这么长时间总结来的经验: 1.坐标系要和题目建的一样,要不就会有各种麻烦. 2.在向前移动过程中碰到其他的机器人也不行,这个题目说啦:a ...
- POJ-2632 Crashing Robots模拟
题目链接: https://vjudge.net/problem/POJ-2632 题目大意: 在一个a×b的仓库里有n个机器人,编号为1到n.现在给出每一个机器人的坐标和它所面朝的方向,以及m条指令 ...
- Crashing Robots(imitate)
Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8124 Accepted: 3528 D ...
- 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 ...
- 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 ...
- poj 2632 Crashing Robots
点击打开链接 Crashing Robots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6655 Accepted: ...
- Poj OpenJudge 百练 2632 Crashing Robots
1.Link: http://poj.org/problem?id=2632 http://bailian.openjudge.cn/practice/2632/ 2.Content: Crashin ...
随机推荐
- SharePoint 2013 PowerShell命令备份还原报错
错误截图: 文字描述: Restore-SPSite : <nativehr>0x80070003</nativehr><nativestack></nati ...
- [javascript svg fill stroke stroke-width rx ry ellipse 属性讲解] svg fill stroke stroke-width ellipse 绘制椭圆属性讲解
<!DOCTYPE html> <html lang='zh-cn'> <head> <title>Insert you title</title ...
- 奇葩问题:This file could not be checked in because the original version of the file on the server was moved or deleted. A new version of this file has been saved to the server, but your check-in comments were not saved
"This file could not be checked in because the original version of the file on the server was m ...
- [Android]使用RecyclerView替代ListView(二)
以下内容为原创,转载请注明: 来自天天博客:http://www.cnblogs.com/tiantianbyconan/p/4242541.html 以前写过一篇“[Android]使用Adapte ...
- Android悬浮窗实现 使用WindowManager
Android悬浮窗实现 使用WindowManager WindowManager介绍 通过Context.getSystemService(Context.WINDOW_SERVICE)可以获得 ...
- Android属性之build.prop生成过程分析
Android的build.prop文件是在Android编译时刻收集的各种property(LCD density/语言/编译时间, etc.),编译完成之后,文件生成在out/target/pro ...
- iOS---检测系统通知开关状态
if (iOS8) { //iOS8以上包含iOS8 if ([[UIApplication sharedApplication] currentUserNotificationSettings].t ...
- 【读书笔记】iOS网络-使用Game Kit实现设备间通信
Apple的Game Kit框架可以实现没有网络状况下的设备与设备之间的通信,这包括没有蜂窝服务,无法访问Wi-Fi基础设施以及无法访问局域网或Internet等情况.比如在丛林深处,高速公路上或是建 ...
- mac PHP配置
apache默认路径配置方法 apache的配置 apache已经自带了,只需如下三个命令就可以了. 开启apache服务 sudo apachectl start 停止apache服务 sudo ...
- Git 的 .gitignore 配置
.gitignore 配置文件用于配置不需要加入版本管理的文件,配置好该文件可以为我们的版本管理带来很大的便利,以下是个人对于配置 .gitignore 的一些心得. 1.配置语法: 以斜杠“/”开头 ...