poj 4048 wa了n次 印象深刻

#include<stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <iostream>
#define max 5000
using namespace std;
const double eps = 1e-8;
struct point{double x,y;};
struct line {point a,b;};
line L[max];
point base;
int T,N;
bool dy(double x,double y)	{	return x > y + eps;}	// x > y
bool xy(double x,double y)	{	return x < y - eps;}	// x < y
bool dyd(double x,double y)	{ 	return x > y - eps;}	// x >= y
bool xyd(double x,double y)	{	return x < y + eps;} 	// x <= y
bool dd(double x,double y) 	{	return fabs( x - y ) < eps;}  // x == y
double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向
{
	return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
double dis(point p1,point p2)
{
  return((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
point intersection(line l1,line l2)
{
  point ans = l1.a;
  double t = ((l1.a.x - l2.a.x)*(l2.a.y - l2.b.y) - (l1.a.y - l2.a.y)*(l2.a.x - l2.b.x))/
             ((l1.a.x - l1.b.x)*(l2.a.y - l2.b.y) - (l1.a.y - l1.b.y)*(l2.a.x - l2.b.x));
  ans.x += (l1.b.x - l1.a.x)*t;
  ans.y += (l1.b.y - l1.a.y)*t;
  return ans;
}
double dot(point a,point b,point c)
{
    return (c.x-a.x)*(b.x-a.x)+(c.y-a.y)*(b.y-a.y);
}
bool judge(point p,line l)
{
  line tmp;
  point ans;
  tmp.a=base,tmp.b=p;
    if(dd(crossProduct(base,p,l.a),0)&&dd(crossProduct(base,p,l.b),0))
    return (dyd(dot(base,p,l.a),0)||dyd(dot(base,p,l.b),0));
  if(xyd(crossProduct(base,p,l.a)*crossProduct(base,p,l.b),0))
  {
    ans=intersection(tmp,l);
  return dyd(dot(base,p,ans),0);
  }
    else return false;
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        int num=0;
        scanf("%d",&N);
        for (int i=0;i<N;i++)
        scanf("%lf%lf%lf%lf",&L[i].a.x,&L[i].a.y,&L[i].b.x,&L[i].b.y);
        scanf("%lf%lf",&base.x,&base.y);
        for (int i=0;i<N;i++)
        {
            int tmp1,tmp2;
            tmp1=tmp2=0;
        for (int j=0;j<N;j++)
        {
            tmp1+=(judge(L[i].a,L[j]));
            tmp2+=(judge(L[i].b,L[j]));
        }
        	  if(tmp1>num)
        	  num=tmp1;
        	  if(tmp2>num)
        	  num=tmp2;
        }
        printf("%d\n",num);

    }
return 0;
}

枚举每个端点和原点组成的射线与其他线段是否相交，若相交还需判断交点是否在射线上，用点积判断

开始怕超时用极角排序，但更麻烦，每个极角对应一个极角范围，逆时针旋转扫描，用正负符号表示线段的开始和结束极角，但是会出现极角范围大于180度的情况需要特殊处理，实在太麻烦不好搞