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);}}
莫找借口失败，只找理由成功。(不为失败找理由，要为成功找方法
