Intersection——判断两条线段能不能相交

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本文介绍了一种用于判断两线段是否相交的算法，并提供了详细的C++代码实现。该算法通过计算向量积和数量积来确定线段端点的相对位置，从而判断两条线段是否相交。

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Intersection

For given two segments s1 and s2, print “1” if they are intersect, “0” otherwise.

s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.

Input
The entire input looks like:

q (the number of queries)
1st query
2nd query
…
qth query
Each query consists of integer coordinates of end points of s1 and s2 in the following format:

xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3
Output
For each query, print “1” or “0”.

Constraints
1 ≤ q ≤ 1000
-10000 ≤ xpi, ypi ≤ 10000
p0≠p1 and p2≠p3.
Sample Input 1
3
0 0 3 0 1 1 2 -1
0 0 3 0 3 1 3 -1
0 0 3 0 3 -2 5 0
Sample Output 1
1
1
0

#include<bits/stdc++.h>
using namespace std;
class Point
{
public:
    double x,y;
    Point(double _x=0,double _y=0):x(_x),y(_y) {}
    Point operator - (Point p)
    {
        return Point(x-p.x,y-p.y);
    }

    double norm()
    {
        return x*x+y*y;
    }
    double ABS()
    {
        return sqrt(norm());   //俩点间的距离
    }
};
struct Segment
{
    Point p1,p2;
};

double dot(Point a,Point b)//求数量积
{
    return a.x*b.x+a.y*b.y;
}

double cross(Point a,Point b)//求向量积
{
    return a.x*b.y-a.y*b.x;
}

int ccw(Point p0,Point p1,Point p2)//判断三个点相对位置
{
    Point a=p1-p0;
    Point b=p2-p0;
    if(cross(a,b)>0) return 1;//p0,p1,p2成逆时针方向
    if(cross(a,b)<0) return -1;//p0,p1,p2成顺时针方向
    if(dot(a,b)<0) return 2;//p2 p0 p1一次排列在同一直线上
    if(a.norm()<b.norm()) return -2;// p0 p1 p2一次排列在同一直线上
    return 0;//p2在线段p0p1上

}
bool intersect(Point p1,Point p2,Point p3,Point p4)
{
    return (ccw(p1,p2,p3)*ccw(p1,p2,p4)<=0&&ccw(p3,p4,p1)*ccw(p3,p4,p2)<=0);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        double x0,y0,x1,y1,x2,y2,x3,y3;
        scanf("%lf %lf %lf %lf %lf %lf %lf %lf",&x0,&y0,&x1,&y1,&x2,&y2,&x3,&y3);
        Point p0(x0,y0),p1(x1,y1),p2(x2,y2),p3(x3,y3);
        if(intersect(p0,p1,p2,p3)) printf("1\n");
        else printf("0\n");
    }

    return 0;
}