hdu 4643 GSM 计算几何 - 点线关系

/*
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;
}