“玲珑杯”ACM比赛 Round #4 G -- See car

G -- See car

Time Limit：2s Memory Limit：64MByt

DESCRIPTION

You are the god of cars, standing at (a, b) point.There are some cars at point $(x_i,y_i)$. If lots of cars and you are in one line, you can only see the car that is nearest to yourself. How many cars can you see?
It is guaranteed that $x_i>a$ and $y_i>b$.

INPUT

There are multiple test cases.The first line is a number T (T $\le 11$), which means the number of cases.For each case, first line has three integers $a,b,n(-{10}^9\le a,b\le {10}^9,0\le n\le {10}^5)$.next $n$ lines, each line contains two integer $(x_i,y_i)(-{10}^9\le x_i,y_i\le {10}^9)$, which means the position of car.

OUTPUT

one line --- the number of car that you can see.

SAMPLE INPUT

2

0 0 3

1 1 

2 2

3 3

0 0 4

1 1

2 2

2 3

4 6

SAMPLE OUTPUT

1

2

题意：给出自己的所在点，再给出其他车的所在坐标，问你从自己所在位置最多能看到多少辆车，每一个方向只能看到这个方向的第一辆车

#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1e5 + 10;
set s;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int a,b,q;
		int count = 0;
		s.clear();
		scanf("%d%d%d",&a,&b,&q);
		while(q--)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			double d = (double)(y - b) / (double)(x - a); // 求斜率，用斜率表示方向
			if(s.find(d) != s.end())
			   continue;
			else
			{
				s.insert(d);
				count++;
			}			
		}
		printf("%d\n",count);
	}
	return 0;
}