發(fā)現由于角的度數和邊的長度有限制，那倆圓如果放得下的話，必然是塞在兩個角里。

于是預處理n個圓心的位置（注意要判斷那個圓會不會和其他的邊界相交），然后n^2枚舉倆角即可。

#include<cstdio> #include<cmath> using namespace std; #define EPS 0.00000001 struct Point{double x,y;double length(){return sqrt(x*x+y*y);} }a[2010]; typedef Point Vector; Vector unit(Vector v){double l=v.length();return (Vector){v.x/l,v.y/l}; } Vector operator - (const Point &a,const Point &b){return (Vector){a.x-b.x,a.y-b.y}; } Vector operator + (const Point &a,const Point &b){return (Vector){a.x+b.x,a.y+b.y}; } double dot(const Vector &a,const Vector &b){return a.x*b.x+a.y*b.y; } double Cross(const Vector &a,const Vector &b){return a.x*b.y-a.y*b.x; } Vector operator * (const double &K,const Vector &v){return (Vector){K*v.x,K*v.y}; } int n; Point calc(int I){double jiao=acos(dot(a[I+1]-a[I],a[I-1]-a[I])/(a[I+1]-a[I]).length()/(a[I-1]-a[I]).length());double d=4000.0/tan(jiao/2.0);double l=4000.0/sin(jiao/2.0);Point p1=a[I]+d*unit(a[I+1]-a[I]);Point p2=a[I]+d*unit(a[I-1]-a[I]);Point M=(Point){(p1.x+p2.x)/2.0,(p1.y+p2.y)/2.0};return a[I]+l*unit(M-a[I]); } double DisToSegment(Point P,Point A,Point B) {Vector v1=B-A,v2=P-A,v3=P-B;if(dot(v1,v2)<EPS) return v2.length();else if(dot(v1,v3)>EPS) return v3.length();else return fabs(Cross(v1,v2))/v1.length(); } Point yuanxins[2010]; bool can[2010]; int main(){ // freopen("k.in","r",stdin);scanf("%d",&n);for(int i=1;i<=n;++i){scanf("%lf%lf",&a[i].x,&a[i].y);}a[++n]=a[1];for(int i=n;i>=1;--i){a[i+1]=a[i];}a[1]=a[n];++n;for(int i=2;i<n;++i){yuanxins[i]=calc(i);can[i]=1;for(int j=2;j<n;++j){ // double tmp=DisToSegment(yuanxins[i],a[j],a[j+1]);if(DisToSegment(yuanxins[i],a[j],a[j+1])-4000.0<-EPS){can[i]=0;break;}}}for(int i=2;i<n;++i){if(Cross(a[i+1]-a[i],a[i]-a[i-1])>EPS && can[i]){Point A=yuanxins[i];for(int j=i+1;j<n;++j){if(Cross(a[j+1]-a[j],a[j]-a[j-1])>EPS && can[j]){Point B=yuanxins[j]; // double tmp=(B-A).length();if((B-A).length()-8000.0>-EPS){printf("%.8f %.8f\n%.8f %.8f\n",A.x,A.y,B.x,B.y);return 0;}}}}}puts("impossible");return 0; }

轉載于:https://www.cnblogs.com/autsky-jadek/p/7198403.html

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