“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<cstdio>#include<cstring>#include<set>#include<vector>#include<iostream>#include<algorithm>using namespace std;const int maxn = 1e5 + 10;set<double> 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;}