判直线和线段相交——poj3304
/* 线段和直线非严格相交:线段两点和直线的叉积 */ #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; #define N 405 #define db double const db eps=1e-8; int sign(db k){ if (k>eps) return 1; else if (k<-eps) return -1; return 0; } int cmp(db k1,db k2){return sign(k1-k2);} struct point{ db x,y; point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};} point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};} point operator * (db k1) const{return (point){x*k1,y*k1};} point operator / (db k1) const{return (point){x/k1,y/k1};} int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;} // 逆时针旋转 point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};} point turn90(){return (point){-y,x};} bool operator < (const point k1) const{ int a=cmp(x,k1.x); if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1; } db abs(){return sqrt(x*x+y*y);} db abs2(){return x*x+y*y;} db dis(point k1){return ((*this)-k1).abs();} point unit(){db w=abs(); return (point){x/w,y/w};} void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;} void print(){printf("%.11lf %.11lf\n",x,y);} db getw(){return atan2(y,x);} point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);} int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);} }; struct line{ point p[2]; line(){} line(point k1,point k2){p[0]=k1; p[1]=k2;} point & operator [] (int k){return p[k];} }; db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;} db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} point p[N]; line lines[N]; int n; int checkLS(point k1,point k2,point k3,point k4){//判 L(k1,k2) 和 S(k3,k4)交点 return sign(cross(k3-k1,k2-k1))*sign(cross(k4-k1,k2-k1))<=0; } int check(point k1,point k2){ if(sign(k1.dis(k2))==0)return false; for(int i=1;i<=n;i++) if(!checkLS(k1,k2,lines[i][0],lines[i][1]))return 0; return 1; } int main(){ int t;cin>>t; while(t--){ scanf("%d",&n); for(int i=1;i<=n;i++){ point k1,k2; scanf("%lf%lf%lf%lf",&k1.x,&k1.y,&k2.x,&k2.y); lines[i]=line(k1,k2); } int flag=0; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){ if(check(lines[i][0],lines[j][0]) || check(lines[i][0],lines[j][1]) || check(lines[i][1],lines[j][1]) || check(lines[i][1],lines[j][0])) flag=1; } if(flag)puts("Yes!"); else puts("No!"); } } /* 2 3 0 0 2 0 0 1 2 1 1 2 1 1 */