Crashing Robots
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 7799   Accepted: 3388

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 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

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 简单的模拟
应用时:15min
实际用时:1h40min
问题:碰撞的先后次序
#include<cstdio>
#include <cstring>
#include <algorithm>
using namespace std; int A,B,n,m;
int robot[][];
char rbuff[];
int dir[];
const int dx[]={,,,-};
const int dy[]={,,-,};
int action[][];
int tx,ty;
int ansr;
bool between(int aim,int gap){
int sx=robot[gap][];
int sy=robot[gap][];
int mingx=min(sx,tx);
int maxgx=max(sx,tx);
int mingy=min(sy,ty);
int maxgy=max(sy,ty);
int ax=robot[aim][];int ay=robot[aim][];
if(ax>=mingx&&ax<=maxgx&&ay>=mingy&&ay<=maxgy){
if(ansr==)ansr=aim;
else {
if(abs(robot[ansr][]-sx)>=abs(ax-sx)&&abs(robot[ansr][]-sy)>=abs(ay-sy)){
ansr=aim;
}
}
return true;
}
return false;
}
void solve(){
ansr=;
for(int i=;i<m;i++){
int rob=action[i][];
int rep=action[i][];
if(action[i][]==){
robot[rob][]=(robot[rob][]+-rep%)%;
}
else if(action[i][]==){
robot[rob][]=(robot[rob][]+rep%)%;
}
else{
bool fl=false;
tx=robot[rob][]+rep*dx[robot[rob][]];
ty=robot[rob][]+rep*dy[robot[rob][]];
for(int j=;j<=n;j++){
if(j==rob)continue;
if(between(j,rob)){
fl=true;
}
}
if(fl){
printf("Robot %d crashes into robot %d\n",rob,ansr);return ;
}
if(tx<||tx>A||ty<||ty>B){
printf("Robot %d crashes into the wall\n",rob);return ;
}
robot[rob][]=tx;
robot[rob][]=ty;
}
}
puts("OK");
} int main(){
#ifndef ONLINE_JUDGE
freopen("output.txt","w",stdout);
#endif // ONLINE_JUDGE
dir['N']=;dir['E']=;dir['S']=;dir['W']=;
dir['L']=;dir['R']=;dir['F']=;
int t;
scanf("%d",&t);
while(t--){
scanf("%d%d%d%d",&A,&B,&n,&m);
for(int i=;i<=n;++i){
scanf("%d%d%s",robot[i],robot[i]+,rbuff);
robot[i][]=dir[rbuff[]];
}
for(int i=;i<m;i++){
scanf("%d%s%d",action[i],rbuff,action[i]+);
action[i][]=dir[rbuff[]];
}
solve();
}
return ;
}

快速切题 poj2632的更多相关文章

  1. 快速切题sgu127. Telephone directory

    127. Telephone directory time limit per test: 0.25 sec. memory limit per test: 4096 KB CIA has decid ...

  2. 快速切题sgu126. Boxes

    126. Boxes time limit per test: 0.25 sec. memory limit per test: 4096 KB There are two boxes. There ...

  3. 快速切题 sgu123. The sum

    123. The sum time limit per test: 0.25 sec. memory limit per test: 4096 KB The Fibonacci sequence of ...

  4. 快速切题 sgu120. Archipelago 计算几何

    120. Archipelago time limit per test: 0.25 sec. memory limit per test: 4096 KB Archipelago Ber-Islan ...

  5. 快速切题 sgu119. Magic Pairs

    119. Magic Pairs time limit per test: 0.5 sec. memory limit per test: 4096 KB “Prove that for any in ...

  6. 快速切题 sgu118. Digital Root 秦九韶公式

    118. Digital Root time limit per test: 0.25 sec. memory limit per test: 4096 KB Let f(n) be a sum of ...

  7. 快速切题 sgu117. Counting 分解质因数

    117. Counting time limit per test: 0.25 sec. memory limit per test: 4096 KB Find amount of numbers f ...

  8. 快速切题 sgu116. Index of super-prime bfs+树思想

    116. Index of super-prime time limit per test: 0.25 sec. memory limit per test: 4096 KB Let P1, P2, ...

  9. 快速切题 sgu115. Calendar 模拟 难度:0

    115. Calendar time limit per test: 0.25 sec. memory limit per test: 4096 KB First year of new millen ...

随机推荐

  1. Eclipse给方法或者类添加自动注释

    自动生成注释: 在团队开发中,注释是必不可少的,为了是自己的注释看起来更加优雅,注释的格式应该统一,我们可以使用Eclipse注释模板自动生成注释. 具体操作如下: 打开注释模板编辑窗口:Window ...

  2. jQuery ajax 添加头部参数跨域

    1.添加HTTP文件头 $.ajax({ url: "http://www.baidu.com", //contentType: "text/html; charset= ...

  3. (转)Spring Boot(一)

    (二期)4.springboot的综合讲解 [课程四]springbo...概念.xmind64.5KB [课程四]spring装配方式.xmind0.2MB [课程四预习]spri...解读.xmi ...

  4. springboot整合mybatis将sql打印到日志(转)

    在前台请求数据的时候,sql语句一直都是打印到控制台的,有一个想法就是想让它打印到日志里,该如何做呢? 见下面的mybatis配置文件: <?xml version="1.0" ...

  5. git源码阅读

    https://github.com/git-for-windows/git/issues/1854 https://github.com/git-for-windows/git/pull/1902/ ...

  6. LeetCode——Maximum Binary Tree

    Question Given an integer array with no duplicates. A maximum tree building on this array is defined ...

  7. js渐隐渐现透明度变化淡入淡出轮播图

    js渐隐渐现透明度变化淡入淡出轮播图.焦点图 一些广告banner展示常见. (附件) <!DOCTYPE html> <html> <head> <meta ...

  8. MySql 插入数据返回数据的Id值

    insert into addeditemgroup(addeditemgroupname) value(') ; select @@IDENTITY  as id; 返回最新的Id:

  9. iBatis的CRUD操作详细总结

    昨天晚上看了一下关于iBatis的一个讲解的视频,讲的和我的这个简单的总结差不多.... 思考了一下还是把主要操作都总结一下吧,当然这里也不是全的,知识简单的CRUD... 首先我觉得持久层的操作主要 ...

  10. [原][osg][osgEarth][粒子特效]关于粒子特效库在osgEarth中,位置摆放问题,跟踪节点移动问题

    首先粒子在地球上位置摆放很简单: //传入的经纬度坐标 osg::Vec3d geoPoint; const SpatialReference* latLong = SpatialReference: ...