找出两条线是否相交

可能重复：
如何检测两条线段的相交位置
确定两条线段是否相交？

给定两条线l1＝((A0，B0(，(A1，B1((和l2＝((A2，B2(，(A3，B3((；Ax、Bx是整数，(Ax、Bx(指定行的开始和结束。

有没有一种算法只使用整数运算来确定l1和l2是否相交？(只需要布尔答案。(

我自己的方法是用定点算法计算交点附近的一个点。然后将溶液(a，b(代入以下方程：

I： abs((A0+a*(A1-A0((-(A2+b*(A3-A2((<公差
II： abs((B0+a*(B1-B0((-(B2+b*(B3-B2((<公差

如果I和II都评估为true，那么我的方法应该返回true。

我的C++-代码：
vec.h：

#ifndef __MY_VECTOR__
#define __MY_VECTOR__
#include <stdarg.h>
template<typename VType, unsigned int dim>
class vec {
private:
    VType data[dim];
public:
    vec(){}
    vec(VType v0, ...){
            data[0] = v0;
            va_list l;
            va_start(l, v0);
            for(unsigned int i=1; i<dim; ++i){
                    data[i] = va_arg(l, VType);
            }
            va_end(l);
    }
    ~vec(){}
    VType& operator[](unsigned int i){
            return data[i];
    }
    VType operator[](unsigned int i) const {
            return data[i];
    }};
    template<typename VType, unsigned int dim, bool doDiv>
    vec<VType, dim> helpArith1(const vec<VType, dim>& A, long delta){
            vec<VType, dim> r(A);
            for(unsigned int i=0; i<dim; ++i){
                    r[i] = doDiv ? (r[i] / delta) : (r[i] * delta);
            }
            return r;
    }
    template<typename VType, unsigned int dim>
    vec<VType, dim> operator*(const vec<VType, dim>& v, long delta) {
        return helpArith1<VType, dim, false>(A, delta);
    }
    template<typename VType, unsigned int dim>
    vec<VType, dim> operator*(long delta, const vec<VType, dim>& v){
        return v * delta;
    }
    template<typename VType,unsigned int dim>
    vec<VType, dim> operator/(const vec<VType, dim>& A, long delta) {
        return helpArith1<VType, dim, true>(A, delta);
    }
    template<typename VType, unsigned int dim, bool doSub>
    vec<VType, dim> helpArith2(const vec<VType, dim>& A, const vec<VType, dim>& B){
        vec<VType, dim> r;
        for(unsigned int i=0; i<dim; ++i){
            r[i] = doSub ? (A[i]-B[i]):(A[i]+B[i]);
        }
        return r;
    }
    template<typename VType, unsigned int dim>
    vec<VType, dim> operator+(const vec<VType, dim>& A, const vec<VType, dim>& B){
        return helpArith2<VType, dim, false>(A, B);
    }
    template<typename VType, unsigned int dim>
    vec<VType, dim> operator-(const vec<VType, dim>& A, const vec<VType, dim>& B){
        return helpArith2<VType, dim, true>(A, B);
    }
    template<typename VType, unsigned int dim>
    bool operator==(const vec<VType, dim>& A, const vec<VType, dim>& B) {
            for(unsigned int i==0; i<dim; ++i){
                if(A[i]!=B[i]){
                            return false;
                    }
            }
            return true;
    }
    template<typename VType, unsigned int dim>
    bool operator!=(const vec<VType, dim>& A, const vec<VType, dim>& B) {
            return !(A==B);
    }
    #endif

line.h:

#ifndef __MY_LINE__
#define __MY_LINE__
#include "vec.h"
unsigned long int ggt(unsigned long int A, unsigned long int B) {
    if(A==0) {
        if(B==0) {
            return 1;
        }
        return B;
    }
    while(B!=0) {
        unsigned long int temp = A % B;
        A = B;
        B = temp;
    }
    return A;
}
#define ABS(n) ( ((n)<0) ? (-n) : (n) )
struct line {
    vec<long int, 2> A, B;
    explicit line(long int iA_0, long int iA_1, long int iB_0, long int iB_1) :
        A(vec<long int, 2>(iA_0<<8, iA_1<<8)),
        B(vec<long int, 2>(iB_0<<8, iB_1<<8)){}
    vec<long int, 2> slope() const{
        vec<long int, 2> temp = A-B;
        if(temp[0]<0) {
            temp[0] = -1 * temp[0];
            temp[1] = -1 * temp[1];
        }
        return temp/ggt(ABS(temp[0]), ABS(temp[1]));
    }
};
bool intersect(line l1, line l2) {
    const long int epsilon = 1<<4;
    vec<long int, 2> sl1 = l1.slope(), sl2 = l2.slope();
    // l2.A + b*sl2 = l1.A + a*sl1
    // <=> l2.A - l1.A = a*sl1 - b*sl2  // = (I, II)^T
    // I': sl2[1] * I; II': sl2[0] * II
    vec<long int, 2> L = l2.A - l1.A, R = sl1;
    L[0] = L[0] * sl2[1];        R[0] = R[0] * sl2[1];
    L[1] = L[1] * sl2[0];        R[1] = R[1] * sl2[0];
    // I' - II'
    long int L_SUB = L[0] - L[1], R_SUB = R[0] - R[1];
    if(ABS(R_SUB) == 0) {
        return ABS(L_SUB) == 0;
    }
    long int temp = ggt(ABS(L_SUB), ABS(R_SUB));
    L_SUB /= temp; R_SUB /= temp;
    // R_SUB * a = L_SUB
    long int a = L_SUB/R_SUB, b = ((l1.A[0] - l2.A[0])*R_SUB + L_SUB * sl1[0])/R_SUB;
    // if the given lines intersect, then {a, b} must be the solution of
    // l2.A - l1.A = a*sl1 - b*sl2
    L = l2.A - l1.A;
    long x = ABS((L[0]- (a*sl1[0]-b*sl2[0]))), y = ABS((L[1]- (a*sl1[1]-b*sl2[1])));
    return x<epsilon && y < epsilon;
}
#endif

main.cpp:

#include "line.h"
int main(){
    line A(0, 0, 6, 0), B(3, 3, 4, -3);
    bool temp = intersect(A, B);
    return 0;
}

(我不确定我的intersect函数是否适用于所有行，但根据我迄今为止使用的测试数据，它返回了正确的结果。(