玲珑杯”ACM比赛 Round #4 G -- See car【Set】

玲珑杯”ACM比赛 Round #4

Start Time：2016-11-05 12:00:00 End Time：2016-11-05 17:00:00 Refresh Time：2016-11-06 21:50:19 Private

G -- See car

Time Limit：2s Memory Limit：64MByte

Submissions：578Solved：218

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

20 0 31 1 2 23 30 0 41 12 22 34 6

SAMPLE OUTPUT

12

SOLUTION

“玲珑杯”ACM比赛 Round #4

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<set>using namespace std;typedef long long LL;const int MAXN=1e5+11;int main(){int T; scanf("%d",&T);while(T--) {int a,b,N; scanf("%d%d%d",&a,&b,&N);set<double> Se;for(int i=0;i<N;++i)  {int x,y; scanf("%d%d",&x,&y);Se.insert((y-b)*1.0/(x-a));}printf("%d\n",Se.size());}return 0;}