LA3890 Most Distant Point from the Sea 二分+半平面交
来源:互联网 发布:乐乎公寓sina 编辑:程序博客网 时间:2024/06/16 21:55
一个凸多边形的岛,求岛上距离海岸最远的距离有多远。
二分距离,然后收缩原直线围成的范围,也就是新的n条直线的半平面交是否为空,不为空就是可行的距离,否则就是不可行的。
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>typedef double type;using namespace std;const double PI=acos(-1.0);const double eps=1e-10;struct Point{ type x,y; Point(){} Point(type a,type b) { x=a; y=b; } void read() { scanf("%lf%lf",&x,&y); } void print() { printf("%.6lf %.6lf",x,y); }};typedef Point Vector;Vector operator + (Vector A,Vector B){ return Vector(A.x+B.x,A.y+B.y);}Vector operator - (Point A,Point B){ return Vector(A.x-B.x,A.y-B.y);}Vector operator * (Vector A,type p){ return Vector(A.x*p,A.y*p);}Vector operator / (Vector A,type p){ return Vector(A.x/p,A.y/p);}bool operator < (const Point &a,const Point &b){ return a.x<b.x || (a.x==b.x && a.y<b.y);}int dcmp(double x){ if (fabs(x)<eps) return 0; else return x<0?-1:1;}bool operator == (const Point& a,const Point b){ return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;}//atan2(x,y) :向量(x,y)的极角,即从x轴正半轴旋转到该向量方向所需要的角度。type Dot(Vector A,Vector B){ return A.x*B.x+A.y*B.y;}type Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}type Length(Vector A){ return sqrt(Dot(A,A));}type Angle(Vector A,Vector B){ return acos(Dot(A,B))/Length(A)/Length(B);}type Area2(Point A,Point B,Point C){ return Cross(B-A,C-A);}Vector Rotate(Vector A,double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A)//单位法线,左转90度,长度归一{ double L=Length(A); return Vector(-A.y/L,A.x/L);}Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){ Vector u=P-Q; double t=Cross(w,u)/Cross(v,w); return P+v*t;}double DistanceToLine(Point P,Point A,Point B){ Vector v1=B-A,v2=P-A; return fabs(Cross(v1,v2))/Length(v1);}double DistanceToSegment(Point P,Point A,Point B){ 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,Point A,Point B)//P在AB上的投影{ Vector v=B-A; return A+v*(Dot(v,P-A)/Dot(v,v));}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){ double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1), c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1); return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;}bool OnSegment(Point p,Point a1,Point a2){ return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0;}double ConvexPolygonArea(Point* p,int n)//多边形面积{ 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;}double PolygonArea(Point* p,int n)//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;}struct Line{ Point p; Vector v; double ang; Line(){}; Line(Point PP,Vector vv) { p=PP; v=vv; ang=atan2(v.y,v.x); } bool operator< (const Line& L)const { return ang<L.ang; } Point point(double t) { return p+v*t; }};struct Circle{ Point c; double r; Circle() { } Circle(Point cc,double rr) { c=cc; r=rr; } Point point(double a) { return Point(c.x+cos(a)*r,c.y+sin(a)*r); }};int getLineCircleIntersection(Line L,Circle C,double& t1,double &t2,vector<Point>& sol){ double a=L.v.x, b=L.p.x-C.c.x, c=L.v.y, d=L.p.y-C.c.y; double e=a*a+c*c,f=2*(a*b+c*d), g=b*b+d*d-C.r*C.r; double delta=f*f-4*e*g;//判别式 if (dcmp(delta)<0) return 0;//相离 if (dcmp(delta)==0) { t1=t2=-f/(2*e); sol.push_back(L.point(t1)); return 1; }//相切 t1=(-f-sqrt(delta))/(2*e); sol.push_back(L.point(t1)); t2=(-f+sqrt(delta))/(2*e); sol.push_back(L.point(t2)); return 2;}double angle(Vector v)//向量极角{ return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point>& sol){ double d = Length(C1.c-C2.c); if (dcmp(d)==0) { if (dcmp(C1.r-C2.r)==0) return -1; return 0; } if (dcmp(C1.r+C2.r-d)<0) return 0; if (dcmp(fabs(C1.r-C2.r)-d)>0) return 0; double a=angle(C2.c-C1.c); double da= acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); Point p1=C1.point(a-da),p2=C1.point(a+da); sol.push_back(p1); if (p1==p2) return 1; sol.push_back(p2); return 2;}int getTangents(Point p,Circle C,Vector* v){ Vector u=C.c-p; double dist=Length(u); if (dist<C.r) return 0; else if (dcmp(dist-C.r)==0) { v[0]=Rotate(u,PI/2); return 1; } else { double ang=asin(C.r/dist); v[0]=Rotate(u,-ang); v[1]=Rotate(u,+ang); return 2; }}int getTangents(Circle A,Circle B,Point* a,Point* b){ int cnt=0; if (A.r<B.r) { swap(A,B); swap(a,b); } int d2=(A.c.x-B.c.x)*(A.c.x-B.c.x)+(A.c.y-B.c.y)*(A.c.y-B.c.y); int rdiff=A.r-B.r; int rsum=A.r+B.r; if (d2<rdiff*rdiff) return 0; double base=atan2(B.c.y-A.c.y,B.c.x-A.c.x); if (d2==0 && A.r==B.r) return -1; if (d2==rdiff*rdiff) { a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++; return 1; } double ang=acos((A.r-B.r)/sqrt((double)d2)); a[cnt]=A.point(base+ang); b[cnt]=B.point(base+ang); cnt++; a[cnt]=A.point(base-ang); b[cnt]=B.point(base-ang); cnt++; if (d2==rsum*rsum) { a[cnt]=A.point(base); b[cnt]=B.point(PI+base); cnt++; } else if (d2>rsum*rsum) { double ang=acos((A.r+B.r)/sqrt((double)d2)); a[cnt]=A.point(base+ang); b[cnt]=B.point(PI+base+ang); cnt++; a[cnt]=A.point(base-ang); b[cnt]=B.point(PI+base-ang); cnt++; } return cnt;}int isPointInpolygon(Point p,Point* poly,int n){ int wn=0; for (int i=0; i<n; i++) { if (OnSegment(p,poly[i],poly[(i+1)%n]) || p==poly[i] || p==poly[(i+1)%n]) return -1; int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[(i+1)%n])); int d1=dcmp(poly[i].y-p.y); int d2=dcmp(poly[(i+1)%n].y-p.y); if (k>0 && d1<=0 && d2>0) wn++; if (k<0 && d2<=0 && d1>0) wn--; } if (wn!=0) return 1; else return 0;}int ConvexHull(Point *p, int n,Point *ch){ 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 (int 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 rotating_calipers(Point *p,int n){ int i,q=1; double ans=0; for (int i=0; i<n-1; i++) { while(Cross(p[q+1]-p[i+1],p[i]-p[i+1])>Cross(p[q]-p[i+1],p[i]-p[i+1])) { q=(q+1)%n; } ans=max(ans,max(Length(p[i]-p[q]),Length(p[i+1]-p[q]))); } return ans*ans;}bool OnLeft(Line L,Point p){ return Cross(L.v,p-L.p)>0;}//两直线交点Point GetIntersection(Line a,Line b){ Vector u=a.p-b.p; double t=Cross(b.v,u)/Cross(a.v,b.v); return a.p+a.v*t;}int HalfplaneIntersection(Line* L,int n,Point* poly){ sort(L,L+n); int first,last; Point *p=new Point[n]; Line *q=new Line[n]; q[first=last=0] = L[0]; for (int i=1; i<n; i++) { while(first<last && !OnLeft(L[i],p[last-1])) last--; while(first<last && !OnLeft(L[i],p[first])) first++; q[++last]=L[i]; if (fabs(Cross(q[last].v,q[last-1].v))<eps) { last--; if (OnLeft(q[last],L[i].p)) q[last]=L[i]; } if (first<last) p[last-1]=GetIntersection(q[last-1],q[last]); } while(first<last && !OnLeft(q[first],p[last-1])) last--; if (last-first<=1) return 0;//空集 p[last]=GetIntersection(q[last],q[first]); //计算首尾两个半平面的交点 int m=0; for (int i=first; i<=last; i++) poly[m++]=p[i]; return m;}Point poly[300];Vector v[300];Vector vm[300];Point ch[300];int tt;int m,n;Line li[300];int main(){// freopen("in.txt","r",stdin); while(~scanf("%d",&n) && n) { for (int i=0; i<n; i++) poly[i].read(); for (int i=0; i<n; i++) { int j=(i+1)%n; v[i]=poly[j]-poly[i]; vm[i]=Normal(v[i]); vm[i]=vm[i]/Length(vm[i]); } double l=0,r=10000; while (fabs(r-l)>1e-7) { double mid=(l+r)/2.0; for (int i=0; i<n; i++) { li[i]=Line(poly[i]+vm[i]*mid,v[i]); } int ck=HalfplaneIntersection(li,n,ch); if (ck>0) { l=mid; } else r=mid; } printf("%.6lf\n",l); } return 0;}
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