POJ 2632：Crashing Robots

标签：poj

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

题意是给定一个坐标地图，给了很多个机器人的坐标和方向，有三种对机器人的操作方式，F是前进，L是向左转，R向右转。看哪一个机器人先撞墙或是撞到其他机器人。

自己需要注意两点：

1.不一定非得到操作机器人的时候才有冲突，可能在最开始的机器人拜访过程中机器人站的位置就已经重复，即冲突了。

2.R向右转，自己定义成为-1，原来自己以为负数除以数的余数会是正数，结果自己too naive了，要考虑将其变为正数。

这个题自己就注意这两点就够了，因为本身就是一个模拟的过程，跟算法也没什么关系。。。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

struct R
{
	int x;
	int y;
	int dir;
}Robot[110];

enum{ E,N,W,S };
enum{ L=1,R=-1};
int map_f[200][200];
int move_x[5]={1,0,-1,0};
int move_y[5]={0,1,0,-1};
int Test,X,Y,Robot_n,Q,i,flag;
string temp;

void solve()
{
	int cal_n,cishu,k;
	string manu;
	cin>>cal_n>>manu>>cishu;

	if(flag==0)
		return;

	if(manu=="F")
	{
		map_f[Robot[cal_n].x][Robot[cal_n].y]=0;

		for(k=1;k<=cishu;k++)
		{
			Robot[cal_n].x = Robot[cal_n].x + move_x[Robot[cal_n].dir];
			Robot[cal_n].y = Robot[cal_n].y + move_y[Robot[cal_n].dir];

			if(Robot[cal_n].x <=0 || Robot[cal_n].y <=0 || Robot[cal_n].x >X || Robot[cal_n].y >Y)
			{
				flag=0;
				cout<<"Robot "<<cal_n<<" crashes into the wall"<<endl;
				break;
			}
			else if(map_f[Robot[cal_n].x][Robot[cal_n].y])
			{
				flag=0;
				cout<<"Robot "<<cal_n<<" crashes into robot "<<map_f[Robot[cal_n].x][Robot[cal_n].y]<<endl;
				break;
			}
		}
		map_f[Robot[cal_n].x][Robot[cal_n].y]=cal_n;
	}
	else if(manu=="L")
	{
		Robot[cal_n].dir = (Robot[cal_n].dir + L*cishu)%4;
	}
	else if(manu=="R")
	{
		Robot[cal_n].dir = (Robot[cal_n].dir + (R*cishu)%4 + 4)%4;
	}
}

int main()
{
	cin>>Test;

	while(Test--)
	{
		flag=1;
		memset(map_f,0,sizeof(map_f));

		cin>>X>>Y;
		cin>>Robot_n>>Q;

		for(i=1;i<=Robot_n;i++)
		{
			cin>>Robot[i].x>>Robot[i].y;
			if(map_f[Robot[i].x][Robot[i].y])
			{
				flag=0;
				cout<<"Robot "<<i<<" crashes into robot "<<map_f[Robot[i].x][Robot[i].y]<<endl;
			}
			else
			{
				map_f[Robot[i].x][Robot[i].y]=i;
			}

			cin>>temp;
			if(temp=="E")
				Robot[i].dir=E;
			else if(temp=="N")
				Robot[i].dir=N;
			else if(temp=="W")
				Robot[i].dir=W;
			else if(temp=="S")
				Robot[i].dir=S;
		}
		for(i=1;i<=Q;i++)
		{
			solve();
		}
		if(flag)
			cout<<"OK"<<endl;
	}
	//system("pause");
	return 0;
}

POJ 2632：Crashing Robots

标签：poj

原文地址：http://blog.csdn.net/u010885899/article/details/47004823