UVa 10969 Sweet Dream 继续圆盘问题
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题目地址:pdf版本
思路:还是统计圆上的每一段弧,和LA2572是很类似的。
我们只需要找到每一段弧是否是可见的就行了,对于每一段弧,必定是一整段的可见状态相同。 (不可能再有交点,因为是枚举的所有交点)。 取出中点判断是否被覆盖就行了。
注意要点:
1 先取出所有交点,对于交点然后再取胡angle+PI;
2 插入0,2*PI
3 算中点的时候要-PI ,还原
4 很奇怪的一点,
if(dcmp(Length(C[k].c-p)-C[k].r)<0)就是wa。
if(dcmp(Length(C[k].c-p)-C[k].r)<=0) 就是ac 等号是怎么也取不到的啊,不知道为什么...
代码:
#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>#include<string>#include<set>const long double eps=5e-13;const long double PI=acos(-1);using namespace std;struct Point{ long double x; long double y; Point(long double x=0,long double y=0):x(x),y(y){} };int dcmp(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bool flag=0;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;}// 求内切圆坐标Circle InscribledCircle(Point p1,Point p2,Point p3){ double a=Length(p2-p3); double b=Length(p3-p1); double c=Length(p1-p2); Point I=(p1*a+p2*b+p3*c)/(a+b+c); return Circle(I,DistanceToLine(I, p1, p2)); }// 外接圆模板怎么证明?Circle CircumscribedCircle(Point p1,Point p2,Point p3){ double Bx=p2.x-p1.x,By=p2.y-p1.y; double Cx=p3.x-p1.x,Cy=p3.y-p1.y; double D=2*(Bx*Cy-By*Cx); double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x; double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y; Point p=Point(cx,cy); return Circle(p,Length(p-p1));}int getCircleCircleIntersection(Circle C1,Circle C2,vector<long double> &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)); sol.push_back(a-da); if(dcmp(da)) { sol.push_back(a+da); return 2; } else return 1; }Circle C[110];int main(){ int T; cin>>T; int n; while(T--) { cin>>n; for(int i=0;i<n;i++) { cin>>C[i].r; C[i].c=read_point(); } long double ans=0; for(int i=0;i<n;i++) { vector<long double> rad; vector<Point> inter; for(int j=0;j<n;j++) { getCircleCircleIntersection(C[i], C[j],inter); } for(int j=0;j<inter.size();j++) { rad.push_back(angle(inter[j]-C[i].c)+PI); } rad.push_back(0); rad.push_back(2*PI); int sz=rad.size(); // for(int j=0;j<sz;j++)// {// while(dcmp(rad[j]-PI)>0) rad[j]-=2*PI;// while(dcmp(rad[j]+PI)<0) rad[j]+=2*PI;// } sort(rad.begin(),rad.end()); // printf("%.3Lf ",rad[j]/PI*180); for(int j=0;j<sz-1;j++) { Point p=C[i].point((rad[j]-PI+rad[j+1]-PI)/2); bool flag=0; for(int k=n-1;k>i;k--) { if(dcmp(Length(C[k].c-p)-C[k].r)<0) { flag=1; break; } } if(!flag) { ans+=(rad[j+1]-rad[j])*C[i].r; //printf("%.3Lf\n",(rad[j+1]-rad[j])*C[i].r); } } } printf("%.3Lf\n",ans); }}
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