cf#21-B-Intersection-计算几何-求直线交点个数

http://codeforces.com/problemset/problem/21/B 

计算几何-给出2条线段求交点个数，AX+BY+C=0，求交点用公式， 

 

 需要先判断是否重合，重合输出-1，有一个坑点是，ABC,可能全为0。AB=0时，若C=0，表示所有点集。若C！=0，表示空集，答案一定为0。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
#include <iostream>
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-6;

struct POINT
{
double x;
double y;
POINT(double a=0, double b=0) { x=a; y=b;}
};

struct LINESEG
{
POINT s;
POINT e;
LINESEG(POINT a, POINT b) { s=a; e=b;}
LINESEG() { }
};struct LINE
{
double a;
double b;
double c;
LINE(double d1=1, double d2=-1, double d3=0) {a=d1; b=d2; c=d3;}
};

bool lineintersect(LINE l1,LINE l2,POINT &p) // 是 L1，L2
{
double d=l1.a*l2.b-l2.a*l1.b;
if(abs(d)<eps) // 不相交
return false;
//p.x = (l2.c*l1.b-l1.c*l2.b)/d;
//p.y = (l2.a*l1.c-l1.a*l2.c)/d;
return true;
}
int equals(double a,double b)
{
return fabs(a-b)<eps;
}
int main()
{
POINT pp;
double a,b,c;
double A,B,C;
scanf("%lf%lf%lf",&a,&b,&c);
LINE l1(a,b,c);
scanf("%lf%lf%lf",&A,&B,&C);
LINE l2(A,B,C);

//重合   xie:-a/b
if (a==0&&b==0&&c!=0)
{
printf("0\n");
return 0;
}if (A==0&&B==0&&C!=0)
{
printf("0\n");
return 0;
}
if (a==0&&b==0&&c==0)
{
printf("-1\n");
return 0;
}
if (A==0&&B==0&&C==0)
{
printf("-1\n");
return 0;
}
if (B==0&&b==0)
{
if (equals(c/a,C/A))
{printf("-1\n");return 0;}

}
else
if (A==0&&a==0)
{
if (equals(c/b,C/B))
{	printf("-1\n");return 0;}
}
else
if (equals(a/a,A/A)&&equals(b/a,B/A)&&equals(c/a,C/A))

{	printf("-1\n");return 0;}

bool ret =lineintersect(l1,l2,pp);
if (ret==true)
printf("1\n");
else
printf("0\n");

return 0;

}