给出平面上两条线段的两个端点,判断这两条线段是否相交(有一个公共点或有部分重合认为相交)。 如果相交,输出"Yes",否则输出"No"。
这道题刘汝佳的的训练指南上有有讲,其中判断端点是否在线段上需要判断四次
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> #include<math.h> const double eps=1e-10; using namespace std; struct Point { double x,y; Point(double x=0,double y=0):x(x),y(y){} }; typedef Point Vector; double Cross(Vector A,Vector B) { return A.x*B.y-A.y*B.x; } Vector operator-(Point A,Point B) { return Vector(A.x-B.x,A.y-B.y); } double Dot(Vector A,Vector B) { return A.x*B.x+A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A,A)); } double Distance(Point P,Point A,Point B) { Vector v1=B-A,v2=P-A; return fabs(Cross(v1,v2))/Length(v1); } int dcmp(double x) { if(fabs(x)<eps) return 0; else return x<0?-1:1; } bool Segment(Point a1,Point a2,Point b1,Point b2) { double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1), c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; } bool onSeg(Point p,Point a1,Point a2) { return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0; } int main() { int T; double xc,yc,R; double x[4],y[4]; scanf("%d",&T); while(T--) { for(int i=0;i<4;i++) scanf("%lf %lf",&x[i],&y[i]); if((x[0]==x[2]&&y[0]==y[2])||(x[0]==x[3]&&y[0]==y[3])||(x[1]==x[2]&&y[1]==y[2])||x[1]==x[3]&&y[1]==y[3]) { printf("Yes\n"); continue; } Point a1(x[0],y[0]); Point a2(x[1],y[1]); Point a3(x[2],y[2]); Point a4(x[3],y[3]); if(Segment(a1,a2,a3,a4)||onSeg(a1,a3,a4)||onSeg(a2,a3,a4)||onSeg(a3,a1,a2)||onSeg(a4,a1,a2)) printf("Yes\n"); else printf("No\n"); } return 0; }