找出两条线是否相交 [英] Find out if 2 lines intersect
问题描述
可能重复:
如何检测两条线段相交的位置?
判断两条线段是否相交?
给定两行 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.
Given are two lines l1=((A0, B0), (A1, B1)) and l2=((A2, B2), (A3, B3)); Ax, Bx are integers and (Ax, Bx) specify the starts and ends of the lines.
是否有一种算法只使用整数运算来确定 l1 和 l2 是否相交?(只需要一个布尔答案.)
Is there a algorithm using only integer arithmetic that determines if l1 and l2 intersect? (Only a Boolean answer is needed.)
我自己的方法是用定点算法计算交点附近的一个点.然后将解 (a, b) 代入以下方程:
My own approach was to compute a point near the intersection point with fixed-point arithmetic. The solution (a, b) was then substituted in the following equations:
I: abs((A0 + a * (A1-A0)) - (A2 + b * (A3-A2))) <宽容
II: abs((B0 + a * (B1-B0)) - (B2 + b * (B3-B2))) <宽容度
I: abs((A0 + a * (A1-A0)) - (A2 + b * (A3-A2))) < tolerance
II: abs((B0 + a * (B1-B0)) - (B2 + b * (B3-B2))) < tolerance
如果 I 和 II 都为真，我的方法应该返回真.
My method should return true if both I and II evaluate to 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，但根据我目前使用的测试数据，它返回了正确的结果.)
(I am not sure if my intersect function works for all lines, but with the test data I used so far it returned with the correct result.)
推荐答案
这是可能的.我们要检查 l1 的两个端点是否在 l2 的不同边，以及 l2 的两个端点是否在 l1 的相对边.
This is possible. We want to check whether both endpoins of l1 are on different sides of l2, and both endpoints of l2 are on opposite sides of l1.
要检查点 (A, B) 在 l1=((A0, B0), (A1, B1)) 的哪一边，我们采取:
To check on which side of l1=((A0, B0), (A1, B1)) a point (A, B) lies, we take:
- 垂直于直线的任意法向量N；一个这样的向量是 (B1-B0, A1-A0)
- 从线的起点到点(A, B)的向量P，即(A-A0, B-B0)
然后我们计算点积:
N · P = (A-A0, B-B0) · (B1-B0, A1-A0) = (A-A0) * (B1-B0) + (B-B0) * (A1-A0)
N · P = (A-A0, B-B0) · (B1-B0, A1-A0) = (A-A0) * (B1-B0) + (B-B0) * (A1-A0)
我们只对符号感兴趣:如果它是正数，则该点位于线的一侧；如果它是负面的，它在另一个.如您所见，不需要浮点运算.
We're only interested in the sign: if it's positive, the point is on one side of the line; if it's negative, it's on the other. As you see, no floating point arithmetic required.
我们可以利用符号相反的数字相乘时总是产生负数这一事实.所以判断两条线段((A0，B0)，(A1，B1))和((A2，B2)，(A3，B3))是否相交的完整表达式是:
We can take advantage of the fact that numbers with opposite signs, when multiplied, always yield negative. So the full expression to determine whether two line segments ((A0, B0), (A1, B1)) and ((A2, B2), (A3, B3)) intersect is:
((A2-A0)*(B1-B0) - (B2-B0)*(A1-A0)) * ((A3-A0)*(B1-B0) - (B3-B0)*(A1-A0)) < 0
&&
((A0-A2)*(B3-B2) - (B0-B2)*(A3-A2)) * ((A1-A2)*(B3-B2) - (B1-B2)*(A3-A2)) < 0
测试代码
一些C++代码来测试上面的计算:
Test Code
Some C++ code to test the above calculation:
#include <iostream>
#include <cstdlib>
struct Point {
int x,y;
};
bool isIntersecting(Point& p1, Point& p2, Point& q1, Point& q2) {
return (((q1.x-p1.x)*(p2.y-p1.y) - (q1.y-p1.y)*(p2.x-p1.x))
* ((q2.x-p1.x)*(p2.y-p1.y) - (q2.y-p1.y)*(p2.x-p1.x)) < 0)
&&
(((p1.x-q1.x)*(q2.y-q1.y) - (p1.y-q1.y)*(q2.x-q1.x))
* ((p2.x-q1.x)*(q2.y-q1.y) - (p2.y-q1.y)*(q2.x-q1.x)) < 0);
}
int main(int argc, char* argv[]) {
if(argc != 9) {
std::cout << "Call as " << argv[0] << " <p1.x> <p1.y> <p2.x> "
<< "<p2.y> <q1.x> <q1.y> <q2.x> <q2.y>" << std::endl;
return -1;
}
Point p1 = {.x = atoi(argv[1]), .y = atoi(argv[2])};
Point p2 = {.x = atoi(argv[3]), .y = atoi(argv[4])};
Point q1 = {.x = atoi(argv[5]), .y = atoi(argv[6])};
Point q2 = {.x = atoi(argv[7]), .y = atoi(argv[8])};
if(isIntersecting(p1,p2,q1,q2)) {
std::cout << "Segments intersect" << std::endl;
return 1;
}
else {
std::cout << "Segments do not intersect" << std::endl;
return 0;
}
}
结果:
$ ./intersection_test 0 0 10 10 0 10 10 0 # example from the comments
Segments intersect
$ ./intersection_test 0 1 2 1 1 2 1 0
Segments intersect
$ ./intersection_test 0 0 0 1 1 1 1 0
Segments do not intersect
$ ./intersection_test 1 1 5 3 3 4 7 2 # q touches but not intersects at p2
Segments do not intersect
$ ./intersection_test 1 1 5 3 3 4 6 2
Segments intersect
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