HDU 3263(Ancient vending machine-井盖问题)

题意：给出一个多边形孔和一个多边形硬币，问你这个硬币能能否从孔中穿过去？

硬币肯定竖着扔，求硬币多边形的最小宽度和多边形孔包含的最长线段长、
多边形的最小宽度就是旋转卡壳时的最小宽度。
然后，有一个结论：
多边形内的最长线段长必然过多边形的两个顶点。

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<complex>#include<stack>#include<map>#include<sstream>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll sqr(ll a){return a*a;}ld sqr(ld a){return a*a;}double sqr(double a){return a*a;}const double eps=1e-10;int dcmp(double x) {    if (fabs(x)<eps) return 0; else return x<0 ? -1 : 1; }ld PI = 3.141592653589793238462643383;class P{public:    double x,y;    P(double x=0,double y=0):x(x),y(y){}    friend ld dis2(P A,P B){return sqr(A.x-B.x)+sqr(A.y-B.y);   }    friend ld Dot(P A,P B) {return A.x*B.x+A.y*B.y; }    friend ld Length(P A) {return sqrt(Dot(A,A)); }    friend ld Angle(P A,P B) {        if (dcmp(Dot(A,A))==0||dcmp(Dot(B,B))==0||dcmp(Dot(A-B,A-B))==0) return 0;        return acos(max((ld)-1.0, min((ld)1.0, Dot(A,B) / Length(A) / Length(B) )) );     }    friend P operator- (P A,P B) { return P(A.x-B.x,A.y-B.y); }    friend P operator+ (P A,P B) { return P(A.x+B.x,A.y+B.y); }    friend P operator* (P A,double p) { return P(A.x*p,A.y*p); }    friend P operator/ (P A,double p) { return P(A.x/p,A.y/p); }    friend bool operator< (const P& a,const P& b) {return dcmp(a.x-b.x)<0 ||(dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)<0 );}}; P read_point() {    P a;    scanf("%lf%lf",&a.x,&a.y);    return a;   } bool operator==(const P& a,const P& b) {    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y) == 0;} typedef P V;double Cross(V A,V B) {return A.x*B.y - A.y*B.x;}double Area2(P A,P B,P C) {return Cross(B-A,C-A);}V Rotate(V A,double rad) {    return V(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));} // A 不是 0向量 V Normal(V A) {     double L = Length(A);    return V(-A.y/L , A.x/L); }namespace complex_G{    typedef complex<double> Point;    //real(p):实部 imag(p):虚部 conj(p):共轭     typedef Point Vector;    double Dot(Vector A,Vector B) {return real(conj(A)*B); }    double Cross(Vector A,Vector B) {return imag(conj(A)*B); }    Vector Rotate(Vector A,double rad) {return A*exp(Point(0,rad)); }}//Cross(v,w)==0(平行)时,不能调这个函数 P GetLineIntersection(P p,V v,P Q,V w){    V u = p-Q;    double t = Cross(w,u)/Cross(v,w);    return p+v*t;}P GetLineIntersectionB(P p,V v,P Q,V w){    return GetLineIntersection(p,v-p,Q,w-Q);}double DistanceToLine(P p,P A,P B) {    V v1 = B-A, v2 = p-A;    return fabs(Cross(v1,v2))/Length(v1);}double DistanceToSegment(P p,P A,P B) {    if (A==B) return Length(p-A);    V 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);}P GetLineProjection(P p,P A,P B) {    V v=B-A;    return A+v*(Dot(v,p-A)/Dot(v,v));}//规范相交-线段相交且交点不在端点 bool SegmentProperIntersection(P a1,P a2,P b1,P 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;}//点在线段上（不包含端点） bool OnSegment(P p,P a1,P a2) {    return dcmp(Cross(a1-p,a2-p)) == 0 && dcmp(Dot(a1-p,a2-p))<0;}double PolygonArea(P *p,int n) {    double area=0;    For(i,n-2) area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;}bool IsCollinear(P A,P B,P C,P D) {    return !dcmp(Cross(A-B,C-D)) && !dcmp(Cross(A-B,C-B)) && !dcmp(DistanceToLine(A,C,D));} /*欧拉公式： V+F-E=2 V-点数 F面数 E边数 */struct C{    P c;    double r,x,y;    C(P c,double r):c(c),r(r),x(c.x),y(c.y){}    P point(double a) {        return P(c.x+cos(a)*r,c.y+sin(a)*r);    }};struct Line{    P p;    V v;    double ang;    Line(){}    Line(P p,V v):p(p),v(v) {ang=atan2(v.y,v.x); }    bool operator<(const Line & L) const {        return ang<L.ang;    }    P point(double a) {        return p+v*a;    }};int getLineCircleIntersection(Line L,C cir,double &t1,double &t2,vector<P> & sol) {    if (dcmp(DistanceToLine(cir.c,L.p,L.p+L.v)-cir.r)==0) {        P A=GetLineProjection(cir.c,L.p,L.p+L.v);        sol.pb(A);         t1 = (A-L.p).x / L.v.x;          return 1;    }    double a = L.v.x, b = L.p.x - cir.c.x, c = L.v.y, d= L.p.y - cir.c.y;    double e = a*a+c*c, f = 2*(a*b + c*d), g = b*b+d*d-cir.r*cir.r;    double delta = f*f - 4*e*g;    if (dcmp(delta)<0) return 0;    else if (dcmp(delta)==0) {        t1 = t2 = -f / (2*e); sol.pb(L.point(t1));        return 1;    }     t1 = (-f - sqrt(delta)) / (2*e); sol.pb(L.point(t1));    t2 = (-f + sqrt(delta)) / (2*e); sol.pb(L.point(t2));    return 2;}double angle(V v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(C C1,C C2,vector<P>& sol) {    double d = Length(C1.c-C2.c);    if (dcmp(d)==0) {        if (dcmp(C1.r - C2.r)==0) return -1; //2圆重合         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));    P p1 = C1.point(a-da), p2 = C1.point(a+da);    sol.pb(p1);    if (p1==p2) return 1;    sol.pb(p2);    return 2; }// Tangents-切线 int getTangents(P p,C c,V* v) {    V 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 getTangentsPoint(P p,C c,P *point) {    V u= c.c-p;    double dist = Length(u);    if (dist<c.r) return 0;    else if (dcmp(dist-c.r)==0) {        point[0]=p;        return 1;    } else {        V v[2];        double ang = asin(c.r / dist);        v[0]=Rotate(u,-ang);point[0]=GetLineProjection(c.c,p,p+v[0]);        v[1]=Rotate(u,ang); point[1]=GetLineProjection(c.c,p,p+v[1]);        return 2;    }}//这个函数假设整数坐标和整数半径//double时要把int改成double int getTangents(C A,C B,P* a,P* 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.y-A.y,B.x-A.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(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(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; }//Circumscribed-外接 C CircumscribedCircle(P p1,P p2,P 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;    P p =P(cx,cy);    return C(p,Length(p1-p));}//Inscribed-内接 C InscribedCircle(P p1,P p2,P p3) {    double a = Length(p2-p3);    double b = Length(p3-p1);    double c = Length(p1-p2);    P p = (p1*a+p2*b+p3*c)/(a+b+c);    return C(p,DistanceToLine(p,p1,p2));}double torad(double deg) {    return deg/180*acos(-1);}//把 角度+-pi 转换到 [0,pi) 上 double radToPositive(double rad) {    if (dcmp(rad)<0) rad=ceil(-rad/PI)*PI+rad;    if (dcmp(rad-PI)>=0) rad-=floor(rad/PI)*PI;     return rad;} double todeg(double rad) {    return rad*180/acos(-1);}//(R,lat,lng)->(x,y,z)void get_coord(double R,double lat,double lng,double &x,double &y,double &z) {    lat=torad(lat);    lng=torad(lng);    x=R*cos(lat)*cos(lng);    y=R*cos(lat)*sin(lng);    z=R*sin(lat);}void print(double a) {    printf("%.6lf",a); }void print(P p) {    printf("(%.6lf,%.6lf)",p.x,p.y);}template<class T>void print(vector<T> v) {    sort(v.begin(),v.end());    putchar('[');    int n=v.size();    Rep(i,n) {        print(v[i]);        if (i<n-1) putchar(',');     }    puts("]");}// 把直线沿v平移// Translation-平移 Line LineTranslation(Line l,V v) {    l.p=l.p+v;    return l;}void CircleThroughAPointAndTangentToALineWithRadius(P p,Line l,double r,vector<P>& sol) {    V e=Normal(l.v);    Line l1=LineTranslation(l,e*r),l2=LineTranslation(l,e*(-r));    double t1,t2;    getLineCircleIntersection(l1,C(p,r),t1,t2,sol);    getLineCircleIntersection(l2,C(p,r),t1,t2,sol);}void CircleTangentToTwoLinesWithRadius(Line l1,Line l2,double r,vector<P>& sol) {    V e1=Normal(l1.v),e2=Normal(l2.v);    Line L1[2]={LineTranslation(l1,e1*r),LineTranslation(l1,e1*(-r))},         L2[2]={LineTranslation(l2,e2*r),LineTranslation(l2,e2*(-r))};    Rep(i,2) Rep(j,2) sol.pb(GetLineIntersection(L1[i].p,L1[i].v,L2[j].p,L2[j].v));}void CircleTangentToTwoDisjointCirclesWithRadius(C c1,C c2,double r,vector<P>& sol) {    c1.r+=r; c2.r+=r;    getCircleCircleIntersection(c1,c2,sol);}//确定4个点能否组成凸多边形，并按顺序（不一定是逆时针）返回 bool ConvexPolygon(P &A,P &B,P &C,P &D) {    if (SegmentProperIntersection(A,C,B,D)) return 1;    swap(B,C);    if (SegmentProperIntersection(A,C,B,D)) return 1;    swap(D,C);    if (SegmentProperIntersection(A,C,B,D)) return 1;    return 0;}bool IsParallel(P A,P B,P C,P D) {    return dcmp(Cross(B-A,D-C))==0;}bool IsPerpendicular(V A,V B) {    return dcmp(Dot(A,B))==0;}//先调用ConvexPolygon 求凸包并确认是否是四边形// Trapezium-梯形 Rhombus-菱形  bool IsTrapezium(P A,P B,P C,P D){    return IsParallel(A,B,C,D)^IsParallel(B,C,A,D);}bool IsParallelogram(P A,P B,P C,P D) {    return IsParallel(A,B,C,D)&&IsParallel(B,C,A,D);}bool IsRhombus(P A,P B,P C,P D) {    return IsParallelogram(A,B,C,D)&&dcmp(Length(B-A)-Length(C-B))==0;}bool IsRectangle(P A,P B,P C,P D) {    return IsParallelogram(A,B,C,D)&&IsPerpendicular(B-A,D-A);}bool IsSquare(P A,P B,P C,P D) {    return IsParallelogram(A,B,C,D)&&IsPerpendicular(B-A,D-A)&&dcmp(Length(B-A)-Length(C-B))==0;}//chord-弦 arc-弧 double ArcDis(double chord,double r) {    return 2*asin(chord/2/r)*r;} typedef vector<P> Polygon ;int isPointInPolygon(P p,Polygon poly) {    int wn=0;    int n=poly.size();    Rep(i,n) {        if (OnSegment(p,poly[i],poly[(i+1)%n])) return -1; //edge        int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));        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; //inside    return 0; //outside}int ConvexHull(P *p,int n,P *ch) {    sort(p,p+n);    int m=0;    Rep(i,n) {        while(m>1 && Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;        ch[m++]=p[i];    }    int k=m;    RepD(i,n-2) {        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;}//把两点式转为一般式 ax+by+c=0 void Two_pointFormToGeneralForm(P A,P B,double &a,double &b,double &c){    a = A.y - B.y;    b = B.x - A.x;    c = Cross(A,B);}//有向直线A->B 切割多边行poly 可能返回单点线段 O(n)Polygon CutPolygon(Polygon poly,P A,P B){    Polygon newpoly;    int n=poly.size();    Rep(i,n) {        P C = poly[i];        P D = poly[(i+1)%n];        if (dcmp(Cross(B-A,C-A))>=0)  newpoly.pb(C);        if (dcmp(Cross(B-A,C-D))) {            P ip = GetLineIntersection(A,B-A,C,D-C);            if (OnSegment(ip,C,D)) newpoly.pb(ip);        }    }    return newpoly;}double PolygonArea(Polygon &p) {    double area=0;    int n=p.size();    For(i,n-2) area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2;} //线上不算 bool OnLeft(Line L,P p) {    return Cross(L.v,p-L.p)>0;} P GetIntersection(Line a,Line b) {    V 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, P* poly) {    sort(L,L+n);    int fi,la;    P *p = new P[n];    Line *q = new Line[n];    q[fi=la=0 ] = L[0];    For(i,n-1) {        while (fi < la && !OnLeft(L[i],p[la-1])) la--;        while (fi < la && !OnLeft(L[i],p[fi])) fi++;        q[++la] = L[i];        if (fabs(Cross(q[la].v, q[la-1].v))<eps) {            la--;            if (OnLeft(q[la],L[i].p)) q[la] = L[i];        }        if (fi<la) p[la-1] = GetIntersection(q[la-1],q[la]);    }    while(fi < la && !OnLeft(q[fi],p[la-1])) la--;    if (la-fi<=1) return 0;     p[la] = GetIntersection(q[la],q[fi]);    int m=0;    Fork(i,fi,la) poly[m++]=p[i];    return m;} // hypot(double x,double y); return sqrt(x*x+y*y)double Jinggai_problems(P *p,int n) {    int q=1;    double ans=INF;    Rep(i,n) {        while(fabs(Area2(p[i],p[(i+1)%n],p[(q+1)%n]))> fabs(Area2(p[i],p[(i+1)%n],p[q] )) )             q=(q+1)%n;        ans = min( ans, (double)DistanceToSegment(p[q],p[i],p[(i+1)%n]) );              }    return ans;}// rotating for rectdouble rec_rotating_calipers(P *p,int n) {    int q=1;    double ans1=1e15,ans2=1e15;    int l=0,r=0;    Rep(i,n) {        while(dcmp(fabs(Area2(p[i],p[(i+1)%n],p[(q+1)%n])) - fabs(Area2(p[i],p[(i+1)%n],p[q] ))) >0  )             q=(q+1)%n;        while (dcmp(Dot(p[(i+1)%n]-p[i],p[(r+1)%n]-p[r]))>0 )  r=(r+1)%n;        if (!i) l=q;        while (dcmp(Dot(p[(i+1)%n]-p[i],p[(l+1)%n]-p[l]))<0 )  l=(l+1)%n;        double d = Length(p[(i+1)%n]-p[i]);        double h = fabs(Area2(p[i],p[(i+1)%n],p[q] ))/d;        double w=(Dot(p[(i+1)%n]-p[i],p[r]-p[i]) - Dot(p[(i+1)%n]-p[i],p[l]-p[i]) ) /d;        ans1=min(ans1,2*(h+w)),ans2=min(ans2,h*w);    }    printf("%.2lf %.2lf\n",ans2,ans1);}//去共线化 void simplify(Polygon& poly) {    Polygon ans;    int n=SI(poly);    Rep(i,n) {        if (dcmp(Cross(poly[i]-poly[(i+1)%n],poly[(i+1)%n]-poly[(i+2)%n]))!=0)            ans.pb(poly[(i+1)%n]);    }    n=SI(ans);    cout<<n<<endl;    Rep(i,n) printf("%.4lf %.4lf\n",ans[i].x,ans[i].y);}int getSegCircleIntersection(Line L,C cir,vector<P> & sol) {    if (dcmp(DistanceToLine(cir.c,L.p,L.p+L.v)-cir.r)==0) {        P A= GetLineProjection(cir.c,L.p,L.p+L.v);        if (OnSegment(A,L.p,L.p+L.v) || L.p==A || L.p+L.v==A  )             sol.pb(A);        return sol.size();    }    double t1,t2;    double a = L.v.x, b = L.p.x - cir.c.x, c = L.v.y, d= L.p.y - cir.c.y;    double e = a*a+c*c, f = 2*(a*b + c*d), g = b*b+d*d-cir.r*cir.r;    double delta = f*f - 4*e*g;    if (dcmp(delta)<0) return 0;    else if (dcmp(delta)==0) {        t1 = -f / (2*e);         if (dcmp(t1)>=0&&dcmp(t1-1)<=0) {            sol.pb(L.point(t1));        }        return sol.size();    }     t1 = (-f - sqrt(delta)) / (2*e); if (dcmp(t1)>=0&&dcmp(t1-1)<=0) sol.pb(L.point(t1));    t2 = (-f + sqrt(delta)) / (2*e); if (dcmp(t2)>=0&&dcmp(t2-1)<=0) sol.pb(L.point(t2));    if(SI(sol)==2 && t1>t2) swap(sol[1],sol[0]);    return sol.size();}int isPointInOrOnCircle(P p,C c) {    return dcmp(Length(p-c.c)-c.r)<=0;}// Triangle(O,A,B) and Circle(O,m)double CircleTriangleArea(P A,P B,double m) {    double ans=0;    C c=C(P(),m); P O=c.c;    if (A==O||B==O) return 0;    bool b = isPointInOrOnCircle(A,c);    bool b2 = isPointInOrOnCircle(B,c);    double opr;    if (dcmp(Area2(O,A,B))>=0) opr=1; else opr=-1;    if (b&&b2) {        ans+=opr*fabs(Area2(A,B,O))/2;     } else if (!b&&!b2){        Line l=Line(A,B-A);        vector<P> sol;        getSegCircleIntersection(l,c,sol);        if (SI(sol)==2) {            ans+=opr*fabs(Area2(sol[0],sol[1],O))/2;            ans+=opr*m*m/2*(Angle(A,sol[0])+Angle(sol[1],B));        } else {            ans+=opr*m*m/2*(Angle(A,B));        }    } else {        Line l=Line(A,B-A);        vector<P> sol;        getSegCircleIntersection(l,c,sol);        if (SI(sol)==2) {            ans+=opr*fabs(Area2(sol[0],sol[1],O))/2;            ans+=opr*m*m/2*(Angle(A,sol[0])+Angle(sol[1],B));        } else if(b) {             ans+=opr*fabs(Area2(sol[0],A,O))/2;            ans+=opr*m*m/2*(Angle(sol[0],B));        } else {            ans+=opr*fabs(Area2(sol[0],B,O))/2;            ans+=opr*m*m/2*(Angle(sol[0],A));        }    }    return ans;}double max_polygon_inside_segment(Polygon p) {    int n=SI(p);    double ans=0;    Rep(i,n) Fork(j,i+1,n-1) {        P A=p[i],B=p[j];        if (A==B) continue;        Polygon poly;        poly.pb(A); poly.pb(B);        Rep(k,n) {            P C=p[k],D=p[(k+1)%n];            if (IsParallel(A,B,C,D)) {                if (IsCollinear(A,B,C,D)) {                    poly.pb(C);                    poly.pb(D);                }            } else {                P t=GetLineIntersectionB(A,B,C,D);                poly.pb(t);            }        }        sort(ALL(poly));        poly.erase(unique(ALL(poly)),poly.end());        int sz=SI(poly);        double l=0; bool fl=0;        For(i,sz-1) {            P m=(poly[i]+poly[i-1])/2;            if (isPointInPolygon(m,p)) {                if(!fl) l=Length(poly[i]-poly[i-1]);                else l+=Length(poly[i]-poly[i-1]);                ans=max(ans,l);                fl=1;            }else fl=0;        }    }    return ans;}P a[100],ch[100];Polygon b;int main(){//  freopen("D.in","r",stdin);//  freopen(".out","w",stdout);    int T=read();    while(T--) {        int k=read();        b.resize(k);        Rep(i,k) b[i]=read_point();        double len2=max_polygon_inside_segment(b);        int n=read();        Rep(i,n) a[i]=read_point();        int m=ConvexHull(a,n,ch);        double len=Jinggai_problems(ch,m);        puts(len<=len2 ? "legal" : "illegal");    }     return 0;}