POJ 2632 较复杂的模拟
来源:互联网 发布:高胡seo博客 编辑:程序博客网 时间:2024/05/19 23:12
Crashing Robots
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9040 Accepted: 3848
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, orF: 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
比较水的一个题,理清思路很快就AC了,中途因为把阀值设为了4搞得wa了一下。 。一怒之下直接改成了4000就AC了。 。顺带一提。 。模拟我写的代码风格都比较。 。丑陋。 。
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;typedef struct{ int loca,sign;}Point;typedef struct{ int x,y;}POint;Point G[150][150];int pp[30];POint loca[150];int m,n;int A,B;POint move[10];bool IsWall(int x,int y){ if(x>=1&&x<=A&&y>=1&&y<=B) return true; return false;}int IsTrue(int x,int y){ if(G[y][x].sign!=-1) return G[y][x].loca; return -1;}int main(){// freopen("data.in","r",stdin); int N; pp['E' - 'A'] = 0; pp['N' - 'A'] = 1; pp['W' - 'A'] = 2; pp['S' - 'A'] = 3; move[0].x = 1;move[0].y = 0; move[1].x = 0;move[1].y = 1; move[2].x = -1;move[2].y = 0; move[3].x = 0;move[3].y = -1; scanf("%d",&N); while(N--){ scanf("%d%d",&A,&B); scanf("%d%d",&n,&m); memset(G,-1,sizeof(G)); memset(loca,0,sizeof(loca)); int i,j; for(i = 1;i<=n;i++){ int xx,yy,kk; char c; scanf("%d %d %c",&xx,&yy,&c); G[yy][xx].sign = pp[c - 'A']; G[yy][xx].loca = i; loca[i].x = xx; loca[i].y = yy; } bool ok = true; for(i = 1;i<=m;i++){ int cur,sum; char c; scanf("%d %c %d",&cur,&c,&sum); if(ok) switch (c){ case 'L': G[loca[cur].y][loca[cur].x].sign = (G[loca[cur].y][loca[cur].x].sign + sum + 4000)%4; break; case 'R': G[loca[cur].y][loca[cur].x].sign = (G[loca[cur].y][loca[cur].x].sign - sum + 4000)%4; break; case 'F': int k = G[loca[cur].y][loca[cur].x].sign; for(j = 0;j<sum;j++){ if(!IsWall(loca[cur].x + move[k].x , loca[cur].y + move[k].y)){ printf("Robot %d crashes into the wall\n",cur); ok = false; break; } int kk = IsTrue(loca[cur].x + move[k].x , loca[cur].y + move[k].y); if(kk!=-1){ printf("Robot %d crashes into robot %d\n",cur,kk); ok = false; break; } G[loca[cur].y + move[k].y][loca[cur].x + move[k].x] = G[loca[cur].y][loca[cur].x]; G[loca[cur].y][loca[cur].x].loca = G[loca[cur].y][loca[cur].x].sign = -1; loca[cur].x = loca[cur].x + move[k].x; loca[cur].y = loca[cur].y + move[k].y; } break; } } if(ok) printf("OK\n"); } return 0;}
- POJ 2632 较复杂的模拟
- zoj 3705 Applications(较复杂的模拟)
- 制作较复杂的电路板
- maven - 较复杂的实例
- 较复杂的sql语句
- POJ 3580 较复杂的Splay区间维护题解题报告
- HDU 4930 Fighting the Landlords --多Trick,较复杂模拟
- 如何应对较复杂的轨迹问题?
- 关于较复杂表达式声明的解析
- Maven入门--较复杂的实例
- 较复杂的ORACLE行列转换
- 较复杂的宏操作符
- Maven入门--较复杂的实例
- Android 较复杂JSON的解析过程
- Maven入门--较复杂的实例
- poj1014(较复杂的背包问题)
- 较复杂的SQL左链接查询
- POJ 3768 Repeater 较复杂 分形 题目
- UIButton, 设置button的背景图片
- 有关报表服务器运行正确,但嵌入到网页中提示因 HTTP 状态 401 失败: Unauthorized。
- 面试时JDBC程序编写简洁版
- UML第二集
- UVA - 11988 Broken Keyboard (a.k.a. Beiju Text)
- POJ 2632 较复杂的模拟
- 中断与异常详解(一)
- 利用运行循环解决NSURLConnection多线程下载的问题(只需要了解)
- Supporting Multiple Screens
- • UFLDL教程练习(exercise)答案(1)
- pjsip项目概要及c语言面向对象方法实现
- Android ORM数据库框架之-greenDao(四)
- git相关操作(简单篇)
- spring web mvc 详解