POJ2653 用叉积判断线段相交

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/liufengwei1/article/details/86563021

第一步：先判断一条线段所在直线与另外一条线段相交

第二步：若上面情况成立，则AC X AD 和 BC X BD要是异号的

#include<cstdio>
#include<cstring>
#define maxl 100010
#define eps 1e-8

struct point
{
	double x,y;
	point(double a=0,double b=0)
	{
		x=a;y=b;
	}
	point operator - (const point &b)const
	{
		return point(x-b.x,y-b.y);
	} 
	double operator * (const point &b)const
	{
		return x*b.x+y*b.y;
	}
	double operator ^ (const point &b)const
	{
		return x*b.y-y*b.x;
	}
};

struct line 
{
	point s,e;
	line(point a=point(),point b=point())
	{
		s=a;e=b;
	}
};

int n;
line a[maxl];
int ans[maxl];
bool top[maxl];

inline void prework()
{
	point p1,p2;
	for(int i=1;i<=n;i++)
	{
		scanf("%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y);
		a[i].s=p1;a[i].e=p2;
		top[i]=true;
	}
}

inline bool cross(line l1,line l2)
{
	line vec=line(l1.s,l1.e-l1.s);
	if((point(l2.s-vec.s)^vec.e)*(vec.e^point(l2.e-vec.s))<0-eps)
		return false;
	if((point(l2.s-l1.s)^point(l2.e-l1.s)) * (point(l2.s-l1.e)^point(l2.e-l1.e))>0+eps)
		return false;
	return true;
}

inline void mainwork()
{
	ans[0]=0;
	for(int i=1;i<=n;i++)
	{	
		for(int j=i+1;j<=n;j++)
		if(cross(a[i],a[j]))
		{
			top[i]=false;
			break;
		}
		if(top[i])
			ans[++ans[0]]=i;
	}
}

inline void print()
{
	printf("Top sticks:");
	for(int i=1;i<=ans[0];i++)
		printf(" %d%c",ans[i],(i==ans[0])?'.':',');
	puts("");
}

int main()
{
	while(~scanf("%d",&n) && n)
	{
		prework();
		mainwork();
		print();
	}
	return 0;
}