/*
hdu 4643 GSM 计算几何 - 点线关系
N个城市，任意两个城市之间都有沿他们之间直线的铁路
M个基站
问从城市A到城市B需要切换几次基站
当从基站a切换到基站b时，切换的地点就是ab的中垂线与铁路的交点(记录由哪两个基站得到的交点，方便切换)处
枚举任意两个基站与铁路的交点，按到城市A的距离排序
求出在城市A时用的基站j，然后开始遍历交点，看从j可以切换到哪个基站(假设是k)，然后再看可以从k可以切换到哪个基站
*/
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
const double eps=1e-11;
struct point
{
double x,y;
}city[55],base[55];
int N,M,K,a,b;
struct node
{
int i,j;
point p;
double dist;
}jiao[3000];
inline bool mo_ee(double x,double y)
{
double ret=x-y;
if(ret<0) ret=-ret;
if(ret<eps) return 1;
return 0;
}
inline bool mo_gg(double x,double y) { return x > y + eps;} // x > y
inline bool mo_ll(double x,double y) { return x < y - eps;} // x < y
inline bool mo_ge(double x,double y) { return x > y - eps;} // x >= y
inline bool mo_le(double x,double y) { return x < y + eps;} // x <= y
inline double min(double a,double b)
{
if(a<b) return a;
return b;
}
inline double max(double a,double b)
{
if(a>b) return a;
return b;
}
point getxiang(point xiang)//求法向量
{
point a;
if(mo_ee(xiang.x,0))
{
a.x=1;
a.y=0;
return a;
}else if(mo_ee(xiang.y,0))
{
a.x=0;
a.y=1;
return a;
}else
{
a.x=1;
a.y=-1.0*xiang.x/xiang.y;
return a;
}
}
inline double mo_distance(point p1,point p2)
{
return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
point mo_intersection(point u1,point u2,point v1,point v2)//两个直线的交点
{
point ret=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))
/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ret.x+=(u2.x-u1.x)*t;
ret.y+=(u2.y-u1.y)*t;
return ret;
}
int segjiao(point &ji,point a,point b,point c,point d)//直线(中垂线)与线段(铁路)的交点
{
ji=mo_intersection(a,b,c,d);
if(mo_ll(ji.x,min(c.x,d.x))) return 0;
if(mo_ll(ji.y,min(c.y,d.y))) return 0;
if(mo_gg(ji.x,max(c.x,d.x))) return 0;
if(mo_gg(ji.y,max(c.y,d.y))) return 0;
return 1;
}
node jiaojiao(int a,int b,int ii,int jj)//求交点
{
node cur;
point xiang1,xiang2;
xiang1.x=city[b].x-city[a].x,xiang1.y=city[b].y-city[a].y;
xiang2.x=base[jj].x-base[ii].x,xiang2.y=base[jj].y-base[ii].y;
if(mo_ee(xiang1.x*xiang2.x,xiang1.y*xiang2.y))
{
cur.i=-1;
return cur;
}
xiang2=getxiang(xiang2);
point zhong;
zhong.x=(base[ii].x+base[jj].x)/2,zhong.y=(base[ii].y+base[jj].y)/2;
point zhongxia;
zhongxia.x=zhong.x+xiang2.x;
zhongxia.y=zhong.y+xiang2.y;
point jiaodian;
int jjao=segjiao(jiaodian,zhong,zhongxia,city[a],city[b]);
if(jjao==0)
{
cur.i=-1;
return cur;
}
cur.p=jiaodian;
cur.i=ii;
cur.j=jj;
cur.dist=mo_distance(jiaodian,city[a]);
return cur;
}
bool cmp(const node &aa,const node &bb)//按交点到城市A的距离排序
{
if(mo_ee(aa.dist,bb.dist))
{
point temp;
temp.x=(base[aa.i].x+base[aa.j].x)/2;
temp.y=(base[aa.i].y+base[aa.j].y)/2;
double dist1=mo_distance(aa.p,temp);
temp.x=(base[bb.i].x+base[bb.j].x)/2;
temp.y=(base[bb.i].y+base[bb.j].y)/2;
double dist2=mo_distance(bb.p,temp);
return mo_ll(dist1,dist2);
}
return mo_ll(aa.dist,bb.dist);
}
inline int nextno(int j,int no)//切换
{
if(jiao[j].i==no) return jiao[j].j;
if(jiao[j].j==no) return jiao[j].i;
return no;
}
int main()
{
int i,j,ncase;
// freopen("1001.in","r",stdin);
// freopen("1001.out.2","w",stdout);
while(scanf("%d%d",&N,&M)!=EOF)
{
for(i=1;i<=N;++i)
{
scanf("%lf%lf",&city[i].x,&city[i].y);
}
for(i=1;i<=M;++i)
{
scanf("%lf%lf",&base[i].x,&base[i].y);
}
scanf("%d",&K);
for(ncase=0;ncase<K;++ncase)
{
int yong=0;
scanf("%d%d",&a,&b);
for(i=1;i<=M;++i)
{
for(j=i+1;j<=M;++j)
{
if(i==j) continue;
node cur=jiaojiao(a,b,i,j);//求任意两个基站与铁路的交点
if(cur.i<0)//无交点
{
continue;
}else
{
jiao[yong++]=cur;
}
}
}
sort(jiao,jiao+yong,cmp);//按到城市a的距离排序
double mindist=-1;
int minno;
for(i=1;i<=M;++i)//求城市a用的基站
{
double disttemp=sqrt((city[a].x-base[i].x)*(city[a].x-base[i].x)+(city[a].y-base[i].y)*(city[a].y-base[i].y));
if(mindist<0||mo_ll(disttemp,mindist))
{
mindist=disttemp;
minno=i;
}
}
int dang=minno,ret=0;
for(i=0;i<yong;++i)//判断切换次数
{
int xinno=nextno(i,dang);
if(xinno!=dang)
{
dang=xinno;
ret++;
}
}
printf("%d\n",ret);
}
}
return 0;
}