POJ2632 Crashing Robots
2015-09-10 15:53
330 查看
Crashing Robots
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
Sample Output
题目大意:有A*B这么大的仓库,有N个机器人,每个机器人占一个格子,并且有一个初始方向,之后有M条指令,格式为
< robot #> < action> < repeat> 让#号机器人执行<
action> 指令< repeat> 次,问是否会出现机器人相撞,或者撞墙。(不会出现两个机器人同时行动)
思路: 老老实实模拟,注意方向位置和大小写(Wa了7次),还有给的图并不是常见的X 0 to n, Y 0 to n注意存图的方式。
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8576 | Accepted: 3703 |
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
题目大意:有A*B这么大的仓库,有N个机器人,每个机器人占一个格子,并且有一个初始方向,之后有M条指令,格式为
< robot #> < action> < repeat> 让#号机器人执行<
action> 指令< repeat> 次,问是否会出现机器人相撞,或者撞墙。(不会出现两个机器人同时行动)
思路: 老老实实模拟,注意方向位置和大小写(Wa了7次),还有给的图并不是常见的X 0 to n, Y 0 to n注意存图的方式。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> using namespace std; #define inf 0xffffff #define maxn 105 int dis[4][2] = {0,1,-1,0,0,-1,1,0}; struct Robot { int x,y,f; }R[maxn]; struct Order { int a,b; char c; }O[maxn]; int Map[maxn][maxn],m,n,a,b,r1,r2; bool flag; void fun() { for(int i = 0; i < m; i++){ if(O[i].c == 'F'){ int xx = R[O[i].a].x; int yy = R[O[i].a].y; int ff = R[O[i].a].f; Map[xx][yy] = 0; for(int j = 0; j < O[i].b; j++){ xx += dis[ff][0]; yy += dis[ff][1]; if(Map[xx][yy] > 0) {flag = true; r1 = O[i].a,r2 = Map[xx][yy];break;} else if( xx < 1 || xx > a || yy < 1 || yy > b) {flag = true; r1 = O[i].a,r2 = 0;break;} } if(flag) break; R[O[i].a].x = xx; R[O[i].a].y = yy; Map[xx][yy] = O[i].a; } else if(O[i].c == 'R') R[O[i].a].f = (R[O[i].a].f + 3 * O[i].b) % 4; else if(O[i].c == 'L') R[O[i].a].f = (R[O[i].a].f + O[i].b) % 4; } } int main() { int k; char dir; scanf("%d",&k); getchar(); while(k--){ flag = false; memset(Map,0,sizeof(Map)); scanf("%d %d",&a, &b); getchar(); scanf("%d %d",&n, &m); getchar(); for(int i = 1; i <= n; i++){ scanf("%d %d %c",&R[i].x,&R[i].y,&dir); getchar(); if(dir == 'N') R[i].f = 0; else if(dir == 'W') R[i].f = 1; else if(dir == 'S') R[i].f = 2; else R[i].f = 3; Map[R[i].x][R[i].y] = i; } for(int i = 0; i < m; i++) {scanf("%d %c %d",&O[i].a,&O[i].c,&O[i].b); getchar();} fun(); if(flag){ if(r2 == 0) printf("Robot %d crashes into the wall\n",r1); else printf("Robot %d crashes into robot %d\n",r1,r2); } else cout<<"OK"<<endl; } return 0; }
相关文章推荐
- 0909 对编译原理的理解
- 16、先天八卦与后天八卦各自有什么用途?
- NSarray 赋值 拷贝 等问题记录
- iOS项目更新之升级Xcode7 & iOS9
- 文章标题
- datetime与smalldatetime之间的区别
- [Android Studio 权威教程]打包、生成jks密钥、签名Apk、多渠道打包
- "ORA-01012: not logged on"以及"Connected to an idle instance."解决思路
- 数据结构(链表)
- Android AlertDialog
- 0909关于编译原理的初印象
- 同时新建AD账号、邮箱、添加通讯组的powershell脚本
- 通过使用的新的浏览器API过滤多余节点
- org.apache.hadoop.hbase.mapreduce.Driver 导入数据到HBASE table
- Extjs6 设置Store、Ajax、form的请求方式(GET、POST)
- asp.net WebForm程序删除.designer.cs文件之后的故事
- arm交叉编译器gnueabi、none-eabi、arm-eabi、gnueabihf、gnueabi区别
- 发现apache的mod_python存在一个BUG
- Using $this when not in object context in
- Service Provider