Crashing Robots
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8124   Accepted: 3528

Description

In a modernized warehouse, robots are used to fetch the goods. Careful planning is needed to ensure that the robots reach their destinations without crashing into each other. Of course, all warehouses are rectangular, and all robots occupy a circular floor space with a diameter of 1 meter. Assume there are N robots, numbered from 1 through N. You will get to know the position and orientation of each robot, and all the instructions, which are carefully (and mindlessly) followed by the robots. Instructions are processed in the order they come. No two robots move simultaneously; a robot always completes its move before the next one starts moving.  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 first line of input is K, the number of test cases. Each test case starts with one line consisting of two integers, 1 <= A, B <= 100, giving the size of the warehouse in meters. A is the length in the EW-direction, and B in the NS-direction.  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 warehouseFinally 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

Output one line for each test case:

  • 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

 #include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
using namespace std;
struct robot
{
int x , y ;
int dir ;
}a[]; struct instruction
{
int num ;
char act ;
int rep ;
}ins[]; int EW , NS ;
int n , m ;
bool flag ;
int map[][]; void crash (int No)
{
int k = ins[No].rep ;
int t = ins[No].num ;
map[a[t].x][a[t].y] = ;
switch (a[t].dir)
{
case :
for (int i = ; i < k && a[t].x && a[t].y ; i++)
if (map[a[t].x][++a[t].y]) {
flag = ;
printf ("Robot %d crashes into robot %d\n" , t , map[a[t].x][a[t].y]) ;
break ;
}
break ;
case :
for (int i = ; i < k && a[t].x && a[t].y ; i++)
if (map[++a[t].x][a[t].y]) {
flag = ;
printf ("Robot %d crashes into robot %d\n" , t , map[a[t].x][a[t].y]) ;
break ;
}
break ;
case :
for (int i = ; i < k && a[t].x && a[t].y ; i++)
if (map[a[t].x][--a[t].y]) {
flag = ;
printf ("Robot %d crashes into robot %d\n" , t , map[a[t].x][a[t].y]) ;
break ;
}
break ;
case :
for (int i = ; i < k && a[t].x && a[t].y; i++)
if (map[--a[t].x][a[t].y]) {
flag = ;
printf ("Robot %d crashes into robot %d\n" , t , map[a[t].x][a[t].y]) ;
break ;
}
break ;
}
map[a[t].x][a[t].y] = t ;
if (a[t].x == || a[t].x >= EW + || a[t].y == || a[t].y >= NS + ) {
printf ("Robot %d crashes into the wall\n" , t) ;
flag = ;
}
}
void solve (int No)
{
int k = ins[No].num ;
switch (ins[No].act)
{
case 'L' : a[k].dir -= ins[No].rep ; a[k].dir %= ; a[k].dir += ; a[k].dir %= ; break ;
case 'R' : a[k].dir += ins[No].rep ; a[k].dir %= ; a[k].dir += ; a[k].dir %= ; break ;
case 'F' : crash (No) ;
}
}
int main ()
{
// freopen ("a.txt" , "r" , stdin) ;
int T ;
scanf ("%d" , &T) ;
char temp ;
while (T--) {
memset (map , , sizeof(map) ) ;
scanf ("%d%d" , &EW , &NS) ;
scanf ("%d%d" , &n , &m) ;
for (int i = ; i <= n ; i++) {
cin >> a[i].x >> a[i].y >> temp ;
// cout << temp <<endl ;
map[a[i].x][a[i].y] = i ;
switch (temp)
{
case 'E' : a[i].dir = ; break ;
case 'S' : a[i].dir = ; break ;
case 'W' : a[i].dir = ; break ;
case 'N' : a[i].dir = ; break ;
}
// printf ("a[%d].dir=%d\n" , i , a[i].dir) ;
}
for (int i = ; i < m ; i++)
cin >> ins[i].num >> ins[i].act >> ins[i].rep ; flag = ;
// printf ("a[1].dir=%d\n" , a[1].dir) ;
for (int i = ; i < m ; i++) {
/* for (int i = 1 ; i <= EW ; i++) {
for (int j = 1 ; j <= NS ; j++) {
printf ("%d " , map[i][j]) ;
}
puts("");
}
puts ("") ;*/
solve (i) ;
/* for (int i = 1 ; i <= EW ; i++) {
for (int j = 1 ; j <= NS ; j++) {
printf ("%d " , map[i][j]) ;
}
puts("");
}
printf ("\n\n\n") ; */
if (flag)
break ;
} if (!flag)
puts ("OK") ;
}
return ;
}

规规矩矩得模拟 robot 一步步走就行了

Crashing Robots(imitate)的更多相关文章

  1. 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 ...

  2. 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 ...

  3. Codeforces Round #625 (Div. 2, based on Technocup 2020 Final Round) A. Contest for Robots(数学)

    题意: n 道题,2 个答题者,已知二者的做题情况,你是受贿裁判,可以给每题指定分值(≥1),求甲乙分数(甲>乙)相差最小时最大分值的最小值. 思路: 统计只有甲或乙做出的题目数. 加一取下整判 ...

  4. POJ 2632 Crashing Robots(较为繁琐的模拟)

    题目链接:http://poj.org/problem?id=2632 题目大意:题意简单,N个机器人在一个A*B的网格上运动,告诉你机器人的起始位置和对它的具体操作,输出结果: 1.Robot i ...

  5. ZOJ 1654 Place the Robots(最大匹配)

    Robert is a famous engineer. One day he was given a task by his boss. The background of the task was ...

  6. [AGC004E] Salvage Robots (DP)

    Description 蛤蟆国的领土我们可以抽象为H*W的笼子,在这片蛤土上,有若干个机器人和一个出口,其余都是空地,每次蛤蟆会要求让所有的机器人向某个方向移动一步,当机器人移动到出口时会被蛤蟆活摘出 ...

  7. 【ZOJ1003】Crashing Balloon(DFS)

    Crashing Balloon Time Limit: 2 Seconds      Memory Limit: 65536 KB On every June 1st, the Children's ...

  8. UVALive 7464 Robots (贪心)

    Robots 题目链接: http://acm.hust.edu.cn/vjudge/contest/127401#problem/K Description http://7xjob4.com1.z ...

  9. UVALive 7464 Robots(模拟)

    7464Robots Write a program to collect data from robots. We are given two sets of robotsX=fX1;:::;Xmg ...

随机推荐

  1. Linux 安装配置Subversion edge

    2014-04-14:修正部分描述.添加JAVA_HOME报错处理步骤.添加配置sudoers 系统:CentOS 5.8 ,6.4 Subversion版本:Subversion Edge 4.0. ...

  2. 一个基于.NET平台的自动化/压力测试系统设计简述

    AutoTest系统设计概述 AutoTest是一个基于.NET平台实现的自动化/压力测试的系统,可独立运行于windows平台下,支持分布式部署,不需要其他配置或编译器的支持.(本质是一个基于协议的 ...

  3. linux下定时任务的使用

    使用方法 执行crontab -e命令会进入一个可编辑界面,在该界面中我们可以制定定时任务,然后保存退出(wq) 格式如下: 由于直接运行编辑命令后只是一个空白界面,不够友好,所以建议使用以下方式来增 ...

  4. Moqui之时间转换

    <script><![CDATA[ if (fromDate == null && thruDate == null && year &&am ...

  5. iOS开发小技巧--设置按钮圆角

    方法一:代码设置 方法二:通过图形化界面

  6. 【CodeForces 606A】A -特别水的题1-Magic Spheres

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102271#problem/A Description Carl is a beginne ...

  7. javascript自动转换大小写

    <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD ...

  8. java 线程---成员变量与局部变量

    关于成员变量与局部变量: 如果一个变量是成员变量,那么多个线程对同一个对象的成员变量进行操作时,他们对该成员变量是彼此影响的(也就是说一个线程对成员变量的改变会影响到另一个线程) . 如果一个变量是局 ...

  9. 安装hadoop2.4.0遇到的问题

    一.执行start-dfs.sh后,datenode没有启动 查看日志如下: 2014-06-18 20:34:59,622 FATAL org.apache.hadoop.hdfs.server.d ...

  10. iphone6 wifi自动断开 tplink

    这样设置就好了,不会自动断开 了