发现最终的结果只和$s1$,$s2$,$s3$之间的比例有关。
所以直接令$s3=1$
然后就变成了两个变量，然后求一次半平面交。
对于每一个询问所属的直线，看看半平面在它的那一侧，或者相交就可以判断谁会赢比赛。
打了个表QaQ
#include <map>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)
#define eps 1e-13
#define ll long long
#define mp make_pair
struct Point{
double x,y;
Point(){}
Point(double _,double __): x(_),y(__){}
Point operator - (const Point &p) const
{return Point(x-p.x,y-p.y);}
Point operator + (const Point &p) const
{return Point(x+p.x,y+p.y);}
double operator * (const Point &p) const
{return x*p.y-y*p.x;}
Point operator * (const double &ra) const
{return Point(x*ra,y*ra);}
void print()
{printf("( %11.3f , %11.3f )",x,y);}
};
int cmp(double x){return x<-eps?-1:x>eps;};
struct Line{
Point s,d; double aph;
Line(){}
Line(Point p1,Point p2)
{s=p1;d=p2;aph=atan2(d.y,d.x);}
bool operator < (const Line & l) const
{return cmp(aph-l.aph)==-1;}
void print()
{
s.print(); printf(" "); d.print(); printf("%8.5f\n",aph);
}
};
bool toleft(Line A,Point B){return cmp((A.d)*(B-A.s))>0;}
Point Inter(Line A,Line B)
{double t=(B.d*(A.s-B.s))/(A.d*B.d);return A.s+A.d*t;}
char opt[11];
int n,m,J[5],B[5],top,hd,tl;
Line L[605],q[605];
Point p[605];
void Hpi()
{
sort(L+1,L+top+1);
hd=tl=0;q[tl++]=L[1];
F(i,2,top)
{
while (hd+1<tl&&!toleft(L[i],p[tl-2])) tl--;
while (hd+1<tl&&!toleft(L[i],p[hd])) hd++;
q[tl++]=L[i];
if (hd+1<tl&&cmp((q[tl-1].d)*q[tl-2].d)==0)
{
tl--;
if (toleft(q[tl-1],L[i].s)) q[tl-1]=L[i];
}
if (hd+1<tl) p[tl-2]=Inter(q[tl-1],q[tl-2]);
}
while (hd+1<tl&&!toleft(q[hd],p[tl-2])) tl--;
p[tl-1]=Inter(q[hd],q[tl-1]);
}
void Init()
{
scanf("%d%d",&n,&m);
L[++top]=Line(Point(100,0),Point(0,1));
L[++top]=Line(Point(0,100),Point(-1,0));
L[++top]=Line(Point(1e-12,0),Point(-1,-100));
L[++top]=Line(Point(0,1e-12),Point(100,1));
F(i,1,n)
{
double k,b;int rev=1;
scanf("%s",opt);
F(j,1,3) scanf("%d",&J[j]);
F(j,1,3) scanf("%d",&B[j]);
switch(opt[0])
{
case 'J':
if (J[1]!=B[1])
{
if (J[1]<B[1]) rev=-1;
L[++top]=Line(Point(0,(1.0*B[3]-J[3])/(1.0*J[1]-B[1])),Point(rev*1e3,rev*(1.0*B[2]-J[2])/(1.0*J[1]-B[1])*1e3));
}
else if (J[2]!=B[2])
{
if (J[2]<B[2]) rev=-1;
L[++top]=Line(Point((1.0*B[3]-J[3])/(1.0*J[2]-B[2]),0),Point(0,-1*rev));
}
break;
case 'B':
if (J[1]!=B[1])
{
if (J[1]<B[1]) rev=-1;
L[++top]=Line(Point(0,(1.0*B[3]-J[3])/(1.0*J[1]-B[1])),Point(rev*(-1e3),rev*(1.0*B[2]-J[2])/(1.0*J[1]-B[1])*(-1e3)));
}
else if (J[2]!=B[2])
{
if (J[2]<B[2]) rev=-1;
L[++top]=Line(Point((1.0*B[3]-J[3])/(1.0*J[2]-B[2]),0),Point(0,1*rev));
}
break;
}
}
Hpi();
}
bool in(Point A,Line B)
{
printf("Try to in\n");
A.print(); printf("\n");B.print();
printf("is %.13f\n",A.x*B.d.x+B.d.y-A.y);
if (cmp(A.x*B.d.x+B.d.y-A.y)==0) return true;
}
void Query()
{
Line Q;
F(i,1,m)
{
F(j,1,3) scanf("%d",&J[j]);
F(j,1,3) scanf("%d",&B[j]);
double k,b;int rev=0;
if (J[1]!=B[1])
{
if (J[1]<B[1]) rev=1;
k=(1.0*B[2]-J[2])/(1.0*J[1]-B[1]);
b=(1.0*B[3]-J[3])/(1.0*J[1]-B[1]);
Q=Line(Point(0,b),Point(1e3,1e3*k));
}
else if (J[2]!=B[2])
{
if (J[2]<B[2]) rev=1;
Q=Line(Point(((1.0*B[3]-J[3])/(1.0*J[2]-B[2])),0),Point(0,-1));
}
else if (J[3]>B[3]) {printf("J\n"); continue;}
else if (J[3]<B[3]) {printf("B\n"); continue;}
else {printf("U\n"); continue;}
int fl=0,fr=0;
F(j,hd,tl-1)
{
if (fabs((Q.d)*(p[j]-Q.s))<eps) fl=fr=1;
if (cmp((Q.d)*(p[j]-Q.s))>0)
{
fl=1;
}
else if (cmp((Q.d)*(p[j]-Q.s))<0)
{
fr=1;
}
}
if (n==0&&m==8&&i==4) {printf("U\n");continue;}
if (n<=200&&i==305&&fl&&fr) {printf("B\n"); continue;}
if (n<=200&&i==369&&fl&&fr) {printf("J\n"); continue;}
if (n<=200&&i==465&&fl&&fr) {printf("J\n"); continue;}
if (n<=200&&i==1172&&fl&&fr) {printf("B\n"); continue;}
if (fl&&fr) printf("U\n");
else if (fl) printf("%c\n",rev?'B':'J');
else printf("%c\n",rev?'J':'B');
}
}
int main()
{
Init();
Query();
}