UVa 11177 Fighting Against a Polygonal Monster 圆与多边形公共面积,二分
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题目地址:pdf版本
如果给定你半径,求公共面积呢,就是之前几题的思路,现在,由于显然 s_common=f(半径) 是关于半径的非减函数,于是二分答案就可以了。
UVa 对long double 的支持也蛮好的~
代码:
#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>const long double eps=1e-10;const long double PI=acos(-1.0);using namespace std;struct Point{ long double x; long double y; Point(long double x=0,long double y=0):x(x),y(y){} void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(long double x) {return (x>eps)-(x<-eps); }int sgn(long double x) {return (x>eps)-(x<-eps); }typedef Point Vector;Vector operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}Vector operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }Vector operator *(Vector A,long double p) { return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,long double p) {return Vector(A.x/p,A.y/p);}ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y<<endl; return out;}//bool operator< (const Point &A,const Point &B) { 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) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}long double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}long double Cross(Vector A,Vector B) {return A.x*B.y-B.x*A.y; }long double Length(Vector A) { return sqrt(Dot(A, A));}long double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}long double Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,long double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {long 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; long double t=Cross(w, u)/Cross(v,w); return P+v*t; }long double DistanceToLine(Point P,Point A,Point B){ Vector v1=P-A; Vector v2=B-A; return fabs(Cross(v1,v2))/Length(v2); }long double DistanceToSegment(Point P,Point A,Point B){ if(A==B) return Length(P-A); Vector v1=B-A; Vector v2=P-A; Vector v3=P-B; if(dcmp(Dot(v1,v2))==-1) return Length(v2); else if(Dot(v1,v3)>0) return Length(v3); else return DistanceToLine(P, A, B); }Point GetLineProjection(Point P,Point A,Point B){ Vector v=B-A; Vector v1=P-A; long double t=Dot(v,v1)/Dot(v,v); return A+v*t;}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){ long double c1=Cross(b1-a1, a2-a1); long double c2=Cross(b2-a1, a2-a1); long double c3=Cross(a1-b1, b2-b1); long double c4=Cross(a2-b1, b2-b1); return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ; }bool OnSegment(Point P,Point A,Point B){ return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;}long double PolygonArea(Point *p,int n){ long 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; }Point read_point(){ Point P; scanf("%Lf%Lf",&P.x,&P.y); return P;}// ---------------与圆有关的--------struct Circle{ Point c; long double r; Circle(Point c=Point(0,0),long double r=0):c(c),r(r) {} Point point(long double a) { return Point(c.x+r*cos(a),c.y+r*sin(a)); } };struct Line{ Point p; Vector v; Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {} Point point(long double t) { return Point(p+v*t); } };int getLineCircleIntersection(Line L,Circle C,long double &t1,long double &t2,vector<Point> &sol){ long double a=L.v.x; long double b=L.p.x-C.c.x; long double c=L.v.y; long double d=L.p.y-C.c.y; long double e=a*a+c*c; long double f=2*(a*b+c*d); long double g=b*b+d*d-C.r*C.r; long double delta=f*f-4*e*g; if(dcmp(delta)<0) return 0; if(dcmp(delta)==0) { t1=t2=-f/(2*e); sol.push_back(L.point(t1)); return 1; } else { t1=(-f-sqrt(delta))/(2*e); t2=(-f+sqrt(delta))/(2*e); sol.push_back(L.point(t1)); sol.push_back(L.point(t2)); return 2; } }// 向量极角公式long double angle(Vector v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){ long double d=Length(C1.c-C2.c); if(dcmp(d)==0) { if(dcmp(C1.r-C2.r)==0) return -1; // 重合 else return 0; // 内含 0 个公共点 } if(dcmp(C1.r+C2.r-d)<0) return 0; // 外离 if(dcmp(fabs(C1.r-C2.r)-d)>0) return 0; // 内含 long double a=angle(C2.c-C1.c); long double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d)); Point p1=C1.point(a-da); Point p2=C1.point(a+da); sol.push_back(p1); if(p1==p2) return 1; // 相切 else { sol.push_back(p2); return 2; }}// 求点到圆的切线int getTangents(Point p,Circle C,Vector *v){ Vector u=C.c-p; long double dist=Length(u); if(dcmp(dist-C.r)<0) return 0; else if(dcmp(dist-C.r)==0) { v[0]=Rotate(u,PI/2); return 1; } else { long double ang=asin(C.r/dist); v[0]=Rotate(u,-ang); v[1]=Rotate(u,+ang); return 2; } }// 求两圆公切线int getTangents(Circle A,Circle B,Point *a,Point *b){ int cnt=0; if(A.r<B.r) { swap(A,B); swap(a, b); // 有时需标记 } long double d=Length(A.c-B.c); long double rdiff=A.r-B.r; long double rsum=A.r+B.r; if(dcmp(d-rdiff)<0) return 0; // 内含 long double base=angle(B.c-A.c); if(dcmp(d)==0&&dcmp(rdiff)==0) return -1 ; // 重合 无穷多条切线 if(dcmp(d-rdiff)==0) // 内切 外公切线 { a[cnt]=A.point(base); b[cnt]=B.point(base); cnt++; return 1; } // 有外公切线的情形 long double ang=acos(rdiff/d); 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(dcmp(d-rsum)==0) // 外切 有内公切线 { a[cnt]=A.point(base); b[cnt]=B.point(base+PI); cnt++; } else if(dcmp(d-rsum)>0) // 外离 又有两条外公切线 { long double ang_in=acos(rsum/d); a[cnt]=A.point(base+ang_in); b[cnt]=B.point(base+ang_in+PI); cnt++; a[cnt]=A.point(base-ang_in); b[cnt]=B.point(base-ang_in+PI); cnt++; } return cnt;}Point Zero=Point(0,0);long double common_area(Circle C,Point A,Point B){ // if(A==B) return 0; if(A==C.c||B==C.c) return 0; long double OA=Length(A-C.c),OB=Length(B-C.c); long double d=DistanceToLine(Zero, A, B); int sg=sgn(Cross(A,B)); if(sg==0) return 0; long double angle=Angle(A,B); if(dcmp(OA-C.r)<=0&&dcmp(OB-C.r)<=0) { return Cross(A,B)/2; } else if(dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)>=0) { return sg*C.r*C.r*angle/2; } else if (dcmp(OA-C.r)>=0&&dcmp(OB-C.r)>=0&&dcmp(d-C.r)<0) { Point prj=GetLineProjection(Zero, A, B); if(OnSegment(prj, A, B)) { vector<Point> p; Line L=Line(A,B-A); long double t1,t2; getLineCircleIntersection(L,C, t1, t2, p); long double s1=0; s1=C.r*C.r*angle/2; long double s2=0; s2=C.r*C.r*Angle(p[0],p[1])/2; s2-=fabs(Cross(p[0],p[1])/2); s1=s1-s2; return sg*s1; } else { return sg*C.r*C.r*angle/2; } } else { if(dcmp(OB-C.r)<0) { Point temp=A; A=B; B=temp; } Point inter_point; long double t1,t2; Line L=Line(A,B-A); vector<Point> inter; getLineCircleIntersection(L, C, t1, t2,inter); if(OnSegment(inter[0], A, B)) inter_point=inter[0]; else { inter_point=inter[1]; } // 两种方法求交点都可以// Point prj=GetLineProjection(Zero, A, B);// // Vector v=B-A;// v=v/Length(v);// long long double mov=sqrt(C.r*C.r-d*d);// // if(OnSegment(prj+v*mov, A, B))// {// inter_point=prj+v*mov;// }// else// {// inter_point=prj+Vector(-v.x,-v.y)*mov;// } long double s=fabs(Cross(inter_point, A)/2); s+=C.r*C.r*Angle(inter_point,B)/2; return s*sg; }}Point p[60];int main(){ int n; long double R; int index=0; while(cin>>n) { if(!n) break; cin>>R; for(int i=0;i<n;i++) p[i]=read_point(); p[n]=p[0]; long double l=0,r=1e10; long double mid=l; while(r-l>1e-6) { mid=l+(r-l)/2; Circle C(Zero,mid); long double ans=0; for(int i=0;i<n;i++) { ans+=common_area(C, p[i], p[i+1]); } if(fabs(ans)>=R) r=mid; else l=mid; } printf("Case %d: %.2Lf\n",++index,mid); } }
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