Description

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

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

Source

题目大意如下：给定几个机器人在一个房间里的坐标和面对的方向（东南西北），再给定几个移动（转几次方向或沿原方向前进其次），判断这些机器人是否会撞到房间墙壁或撞到互相或平安无事。

题解：模拟即可。把方向数组设成按顺时针或逆时针的顺序可缩减代码量。转方向时，注意mod4来减少要转圈数。

代码稍长：

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;//横坐标x代表WE方向,纵坐标y代表NS方向;

//const int mx[]={0,+1,-1,0,0};mov[1]->E,沿x轴向东/右移动+1;mov[2]->W,向西/左移动-1;mov[3]->N,沿x轴不移动;mov[4]->S,沿x轴不移动.

//const int my[]={0,0,0,+1,-1};mov[1]->E,沿y轴不移动;mov[2]->W,沿y轴不移动;mov[3]->N,沿y轴向北/上移动+1;mov[4]->S,沿y轴向南/下移动-1.

//顺序：东南西北;从0开始;

const int mx[]={+1,0,-1,0};

const int my[]={0,-1,0,+1};

int n,m,p,q,mp[101][101];

struct robot{

    int x,y,z;

}rt[101];

struct req{

    int k,rep;

    char pos;

}rq[101];

void Init()

{

    scanf("%d%d%d%d",&n,&m,&p,&q);

    for(int i=1;i<=p;i++){

        char z;

        scanf("%d%d",&rt[i].x,&rt[i].y);

        getchar();

        z=getchar();

        mp[rt[i].x][rt[i].y]=i;

        switch(z){

            case 'E':rt[i].z=0;break;

            case 'S':rt[i].z=1;break;

            case 'W':rt[i].z=2;break;

            case 'N':rt[i].z=3;break;

        }

    }

    for(int i=1;i<=q;i++){

    	scanf("%d",&rq[i].k);

    	getchar();

    	rq[i].pos=getchar();

    	scanf("%d",&rq[i].rep);

    }

}

bool judge(int k)

{

	if(rt[k].x<1||rt[k].x>n||rt[k].y<1||rt[k].y>m){

		printf("Robot %d crashes into the wall\n",k);

		return false;

	}

	if(mp[rt[k].x][rt[k].y]){

		printf("Robot %d crashes into robot %d\n",k,mp[rt[k].x][rt[k].y]);

		return false;

	}

	mp[rt[k].x][rt[k].y]=k;//修改移动后位置数据;

	return true;

}

bool solve()

{

    for(int i=1;i<=q;i++){

    	int k=rq[i].k;

        switch(rq[i].pos){

            case 'F':

            	for(int j=1;j<=rq[i].rep;j++){

            		mp[rt[k].x][rt[k].y]=0;//清空原位置数据;

					rt[k].x+=mx[rt[k].z];

            		rt[k].y+=my[rt[k].z];

            		if(!judge(k))return false;

            	}

				break;

            case 'L':rt[k].z=(rt[k].z-rq[i].rep%4+4)%4;break;

            case 'R':rt[k].z=(rt[k].z+rq[i].rep%4)%4;break;

        }

    }

    return true;

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--){

        memset(mp,0,sizeof(mp));

        Init();

        if(solve())puts("OK");

    }

    return 0;

}