玲珑杯”ACM比赛 Round #4 G -- See car【Set】_玲珑杯 round#4 超时题-优快云博客

本题为“玲珑杯”ACM比赛题目，要求选手编程解决视线内可见车辆数量的问题。输入包括测试案例数量、观测者位置及车辆坐标，输出可见车辆总数。通过计算斜率并利用集合去重实现。

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  (xi,yi)(xi,yi). 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 xi>axi>a and yi>byi>b.

INPUT

   There are multiple test cases.The first line is a number T (T
   
    ≤11 ≤11), which means the number of cases.For each case, first line has three integers 
    
   a,b,n(−109≤a,b≤109,0≤n≤105)a,b,n(−109≤a,b≤109,0≤n≤105).next 
   
   nn lines, each line contains two integer 
    
   (xi,yi)(−109≤xi,yi≤109)(xi,yi)(−109≤xi,yi≤109), 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;
}