模拟——Crashing Robots

Crashing Robots

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7471 Accepted: 3266

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

45 42 21 1 E5 4 W1 F 72 F 75 42 41 1 E5 4 W1 F 32 F 11 L 11 F 35 42 21 1 E5 4 W1 L 961 F 25 42 31 1 E5 4 W1 F 41 L 11 F 20

Sample Output

Robot 1 crashes into the wallRobot 1 crashes into robot 2OKRobot 1 crashes into robot 2

Source

Nordic 2005

一个大模拟：

K组测试数据，A代表列数，B代表行数，N代表机器人的数量，M代表操作的次数。

在N行里，给出每个机器人的初始位置和方向。

在M行里，给出执行操作机器人的代号，要执行的命令，执行该命令的重复次数。

四个方向（W,E,S,N），三种操作：L（左转90度），R（右转90度），F（前进一个方格）。

要求：

按操作执行，如果机器人相遇，则打印相遇，如果机器人撞墙，则打印撞墙，如果顺利执行任务，则打印‘OK’。

思路：

一个结构体，记录当前机器人的位置和状态，按照执行命令开始执行，边执行边判断。

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>#include <stdlib.h>using namespace std;struct node{    int X,Y;    int fx;};struct node bj[109];int main(){    int n,m,i,j,k,N,M,K,A,B,x,y,p,pd;    char ch;    scanf("%d",&K);    while(K--)    {        scanf("%d%d",&A,&B);        memset(bj,0,sizeof(bj));        scanf("%d%d",&N,&M);        for(i = 1; i <= N; i++)        {            scanf("%d %d %c%*c",&x,&y,&ch);            bj[i].X = x;            bj[i].Y = y;            if(ch == 'N')                bj[i].fx = 1;            else if(ch == 'S')                bj[i].fx = 3;            else if(ch == 'E')                bj[i].fx = 2;            else if(ch == 'W')                bj[i].fx = 4;        }        pd = 0;        for(i = 1; i <= M; i++)        {            scanf("%d %c %d",&j,&ch,&p);            if(pd == 1)                continue;            if(ch == 'L')            {                while(p--)                {                    if(bj[j].fx == 1)                        bj[j].fx = 4;                    else                    {                        bj[j].fx--;                    }                }            }            else if(ch == 'R')            {                while(p--)                {                    if(bj[j].fx == 4)                        bj[j].fx = 1;                    else                    {                        bj[j].fx++;                    }                }            }            else if(ch == 'F')            {                while(p--)                {                    if(bj[j].fx == 1)                    {                        bj[j].Y++;                        if(bj[j].Y == B + 1)                        {                            printf("Robot %d crashes into the wall\n",j);                            pd = 1;                            break;                        }                        for(k = 1; k <= N; k++)                        {                            if(k != j)                            {                                if(bj[j].X == bj[k].X && bj[j].Y == bj[k].Y)                                {                                    printf("Robot %d crashes into robot %d\n",j,k);                                    pd = 1;                                    break;                                }                            }                        }                    }                    else if(bj[j].fx == 2)                    {                        bj[j].X++;                        if(bj[j].X == A + 1)                        {                            printf("Robot %d crashes into the wall\n",j);                            pd = 1;                            break;                        }                        for(k = 1; k <= N; k++)                        {                            if(k != j)                            {                                if(bj[j].X == bj[k].X && bj[j].Y == bj[k].Y)                                {                                    printf("Robot %d crashes into robot %d\n",j,k);                                    pd = 1;                                    break;                                }                            }                        }                    }                    else if(bj[j].fx == 3)                    {                        bj[j].Y--;                        if(bj[j].Y == 0)                        {                            printf("Robot %d crashes into the wall\n",j);                            pd = 1;                            break;                        }                        for(k = 1; k <= N; k++)                        {                            if(k != j)                            {                                if(bj[j].X == bj[k].X && bj[j].Y == bj[k].Y)                                {                                    printf("Robot %d crashes into robot %d\n",j,k);                                    pd = 1;                                    break;                                }                            }                        }                    }                    else if(bj[j].fx == 4)                    {                        bj[j].X--;                        if(bj[j].X == 0)                        {                            printf("Robot %d crashes into the wall\n",j);                            pd = 1;                            break;                        }                        for(k = 1; k <= N; k++)                        {                            if(k != j)                            {                                if(bj[j].X == bj[k].X && bj[j].Y == bj[k].Y)                                {                                    printf("Robot %d crashes into robot %d\n",j,k);                                    pd = 1;                                    break;                                }                            }                        }                    }                    if(pd == 1)                        break;                }            }        }        if(pd == 0)            printf("OK\n");    }    return 0;}