题意：容易理解
分析：切换的地点为两个基站所在直线的中垂线与两座城市所在直线的交点。
代码实现：
#include <cstdio> #include <cmath> #include <algorithm> #define maxn 60 #define eps 1e-7 using namespace std; int dcmp(double x) //控制精度 { if(fabs(x)<eps) return 0; else return x<0?-1:1; } double toRad(double deg) //角度转弧度 { return deg/180.0*acos(-1.0); } struct Point { double x,y; Point(){} Point(double x,double y):x(x),y(y) {} void input() { scanf("%lf %lf",&x,&y); } }; typedef Point Vector; Vector operator+( Vector A, Vector B ) //向量加 { return Vector( A.x + B.x, A.y + B.y ); } Vector operator-(Vector A,Vector B) //向量减 { return Vector( A.x - B.x, A.y - B.y ); } Vector operator*( Vector A, double p ) //向量数乘 { return Vector( A.x * p, A.y * p ); } Vector operator/( Vector A, double p ) //向量数除 { return Vector( A.x / p, A.y / p ); } bool operator<(const Point& A, const Point& B ) //两点比较 { return dcmp( A.x - B.x ) < 0 || ( dcmp( A.x - B.x ) == 0 && dcmp( A.y - B.y ) < 0 ); } bool operator==( const Point& a, const Point& b ) //两点相等 { return dcmp( a.x - b.x ) == 0 && dcmp( a.y - b.y ) == 0; } struct Line { Point s,e; Vector v; Line() {} Line(Point s,Point v,int type)://法向量式 s(s),v(v){} Line(Point s,Point e):s(s),e(e)//两点式 {v=e-s;} }; double Dot(Vector A,Vector B)//向量点乘 { return A.x*B.x+A.y*B.y; } double Length(Vector A)//向量模 { return sqrt(Dot(A,A)); } double Angle(Vector A,Vector B)//向量夹角 { return acos(Dot(A,B)/Length(A)/Length(B)); } double Cross(Vector A,Vector B)//向量叉积 { return A.x*B.y-A.y*B.x; } double Area2(Point A,Point B,Point C )//向量有向面积 { return Cross(B-A,C-A); } double Dist(Point A,Point B) { return Length(A-B); } Vector Rotate(Vector A, double rad)//向量逆时针旋转 { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); } Vector Normal(Vector A)//向量单位法向量 { double L=Length(A); return Vector(-A.y/L,A.x/L); } Point GetLineIntersection(Line l1,Line l2)//两直线交点 { Point P=l1.s; Vector v=l1.v; Point Q=l2.s; Vector w=l2.v; Vector u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t; } double DistanceToLine(Point P,Line L)//点到直线的距离 { Point A,B; A=L.s,B=L.e; Vector v1=B-A,v2=P-A; return fabs(Cross(v1,v2))/Length(v1); } double DistanceToSegment(Point P, Line L)//点到线段的距离 { Point A,B; A=L.s,B=L.e; if(A==B) return Length(P-A); Vector v1=B-A,v2=P-A,v3=P-B; if (dcmp(Dot(v1,v2))<0) return Length(v2); else if (dcmp(Dot(v1,v3))>0) return Length(v3); else return fabs(Cross(v1,v2)) / Length(v1); } Point GetLineProjection(Point P,Line L)// 点在直线上的投影 { Point A,B; A=L.s,B=L.e; Vector v=B-A; return A+v*(Dot(v,P-A)/Dot(v,v)); } bool OnSegment(Point p,Line l)//点在线段上包括端点 { Point a1=l.s; Point a2=l.e; return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dist(p,a1)+Dist(p,a2)-Dist(a1,a2))==0; } bool Paralled(Line l1,Line l2)//直线平行 { return dcmp(Cross(l1.e-l1.s,l2.e-l2.s))==0; } bool SegmentProperIntersection(Line l1,Line l2)//线段相交 { if(Paralled(l1,l2)) { return false; } Point t=GetLineIntersection(l1,l2); if(OnSegment(t,l1)) { return true; } return false; } int ConvexHull(Point *p,int n,Point *ch) //求凸包 { sort(p,p+n); int m=0; for ( int i = 0; i < n; ++i ) { while ( m > 1 && Cross( ch[m - 1] - ch[m - 2], p[i] - ch[m - 2] ) <= 0 ) --m; ch[m++] = p[i]; } int k = m; for ( i = n - 2; i >= 0; --i ) { while ( m > k && Cross( ch[m - 1] - ch[m - 2], p[i] - ch[m - 2] ) <= 0 ) --m; ch[m++] = p[i]; } if ( n > 1 ) --m; return m; } double PolygonArea(Point *p,int n) //多边形有向面积 { double area=0; for (int i=1;i<n-1;++i) area+=Cross(p[i]-p[0],p[i+1]-p[0]); return area/2.0; } ///*******************************************************/// int N,M,Q; Point city[maxn],base[maxn]; int solve(int a,int b) { Line train=Line(city[a],city[b]); int res=0; int i,j,k; for(i=0; i<M; i++) { for(j=i+1; j<M; j++) { Line vert=Line((base[i]+base[j])/2,Normal(base[i]-base[j]),1); if(Paralled(train,vert)) { // printf("caonima "); continue; } Point t=GetLineIntersection(train,vert); if(!OnSegment(t,train)) continue; double d1=Dist(t,base[i]); int flag=1; for(k=0;k<M;k++) { if(k==i||k==j) continue; double d2=Dist(t,base[k]); if(dcmp(d2-d1)<0) { flag=0; break; } } if(flag) { res++; } } } return res; } int main() { //freopen("input.txt","r",stdin); int i,j; while(scanf("%d %d",&N,&M)==2) { for(i=0; i<N; i++) { city[i].input(); } for(i=0; i<M; i++) { base[i].input(); } scanf("%d",&Q); while(Q--) { int a,b; scanf("%d %d",&a,&b); a--,b--; printf("%d ",solve(a,b)); } } return 0; }