cf#21-B-Intersection-计算几何-求直线交点个数

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

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