POJ 2632 Crashing Robots（模拟）

Crashing Robots

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8072 Accepted: 3505

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

好吧，我承认自己水了，写完调了两个小时才发现自己理解错题意了。。。

给你n个机器人的坐标、方向和仓库的长和宽，有三种命令 (x 'F' s)指x号机器人沿它的方向走s格，(x 'L' s)指X号机器人的方向向左转s次,(x 'R' s)指X号机器人的方向向右转s次, 在机器人走的过程中，可能会撞墙和撞上其他机器人，这时就要输出,否则输出ok。。。。

#include <iostream>#include <cstring>#include <cstdio>#include <stdlib.h>#include <algorithm>using namespace std;const int N = 110;struct node{    int x,y;    char s;} q[N];int A,B;int n,m;int flag;int map[N][N];int sx,sy;void go(char e){    if(e=='E')    {        sx=1;        sy=0;    }    else if(e=='S')    {        sx=0;        sy=-1;    }    else if(e=='W')    {        sx=-1;        sy=0;    }    else if(e=='N')    {        sx=0;        sy=1;    }}void move(int b,int t){    map[q[b].x][q[b].y]=0;    int bx=q[b].x;    int by=q[b].y;    for(int i=1; i<=t; i++)    {        go(q[b].s);        bx+=sx;        by+=sy;        if(bx==0||bx>A||by==0||by>B)        {            flag=1;            printf("Robot %d crashes into the wall\n",b);            return ;        }        if(map[bx][by])        {            flag=1;            printf("Robot %d crashes into robot %d\n",b,map[bx][by]);            return ;        }        if(i==t)        {            map[bx][by]=b;            q[b].x=bx;            q[b].y=by;            return ;        }    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(map,0,sizeof(map));        scanf("%d%d",&A,&B);        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)        {            int x1,y1;            char s1;            cin>>x1>>y1>>s1;            q[i].x=x1;            q[i].y=y1;            q[i].s=s1;            map[x1][y1]=i;        }        flag=0;        for(int i=1; i<=m; i++)        {            int k,r;            char c;            cin>>k>>c>>r;            if(!flag&&c=='F')                move(k,r);            else if(!flag&&c=='L')            {                char tt=q[k].s;                char str=tt;                r=r%4;                for(int i=1; i<=r; i++)                {                    if(tt=='E')                        str='N';                    else if(tt=='N')                        str='W';                    else if(tt=='W')                        str='S';                    else if(tt=='S')                        str='E';                    tt=str;                }                q[k].s=str;            }            else if(!flag&&c=='R')            {                char tt=q[k].s;                char str=tt;                r=r%4;                for(int i=1; i<=r; i++)                {                    if(tt=='E')                        str='S';                    else if(tt=='S')                        str='W';                    else if(tt=='W')                        str='N';                    else if(tt=='N')                        str='E';                    tt=str;                }                q[k].s=str;            }        }        if(!flag)            printf("OK\n");    }    return 0;}