LA 4728 Squares 旋转卡壳
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题目地址：LA4728
直接白书模板就行
代码：
注意
int n=p.size();
p.push_back(p[0]);
这两句的顺序一定不能反了
#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>const int eps=1e-10;const int PI=acos(-1.0);using namespace std;struct Point{ int x; int y; Point(int x=0,int y=0):x(x),y(y){} void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}};int dcmp(int x) {return (x>eps)-(x<-eps); }int sgn(int 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,int p) { return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,int 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;}int Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}int Cross(Vector A,Vector B) {return A.x*B.y-B.x*A.y; }int Length(Vector A) { return (Dot(A, A));}int Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}int Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,int rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {int 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; int t=Cross(w, u)/Cross(v,w); return P+v*t; }int DistanceToLine(Point P,Point A,Point B){ Vector v1=P-A; Vector v2=B-A; return fabs(Cross(v1,v2))/Length(v2); }int 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; int t=Dot(v,v1)/Dot(v,v); return A+v*t;}// 已针对本题优化bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){ int c1=Cross(b1-a1, a2-a1); int c2=Cross(b2-a1, a2-a1); int c3=Cross(a1-b1, b2-b1); int 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;}int PolygonArea(Point *p,int n){ int 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; int r; Circle(Point c=Point(0,0),int r=0):c(c),r(r) {} Point point(int 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(int t) { return Point(p+v*t); } };int getLineCircleIntersection(Line L,Circle C,int &t1,int &t2,vector<Point> &sol){ int a=L.v.x; int b=L.p.x-C.c.x; int c=L.v.y; int d=L.p.y-C.c.y; int e=a*a+c*c; int f=2*(a*b+c*d); int g=b*b+d*d-C.r*C.r; int 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int angle(Vector v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){ int 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 int a=angle(C2.c-C1.c); int 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; int 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 { int 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 } int d=Length(A.c-B.c); int rdiff=A.r-B.r; int rsum=A.r+B.r; if(dcmp(d-rdiff)<0) return 0; // 内含 int 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 int 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 { int 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;}// 几何算法模板int isPointInPolygon(Point p,Point * poly,int n){ int wn=0; for(int i=0;i<n;i++) { if(OnSegment(p, poly[i], poly[(i+1)%n])) return -1; 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; else return 0; }// Andrew 算法求凸包int ConvexHull(Point *p,int n,Point *ch){ int m=0; sort(p,p+n); n=unique(p, p+n)-p; for(int i=0;i<n;i++) { while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } int k=m; for(int i=n-2;i>=0;i--) { 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; }vector<Point> ConvexHull(vector<Point> p){ sort(p.begin(),p.end()); p.erase(unique(p.begin(), p.end()),p.end()); int n=p.size(); vector<Point> ch(n+1); int m=0; for(int i=0;i<n;i++) { while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--; ch[m++]=p[i]; } int k=m; for(int i=n-2;i>=0;i--) { 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--; ch.resize(m); return ch; }//int max(int a,int b)//{// return a>b?a:b;//}int diameter(vector<Point> poly){ vector<Point> p=ConvexHull(poly); int n=p.size(); p.push_back(p[0]); if(n==1) return 0; else if(n==2) return Length(p[1]-p[0]); int ans=0; for(int u=0,v=1;u<n;u++) { while(1) { int diff=Cross(p[u+1]-p[u],p[v+1]-p[v]); if(dcmp(diff)<=0) { ans=max(ans,Length(p[v]-p[u])); if(dcmp(diff)==0) ans=max(ans,Length(p[v+1]-p[u])); break; } v=(v+1)%n; } } return ans;}int main(){ int T; cin>>T; int n; Point temp; int x,y,w; while(T--) { vector<Point> p; cin>>n; for(int i=0;i<n;i++) { scanf("%d%d%d",&x,&y,&w); p.push_back(Point(x,y)); p.push_back(Point(x+w,y)); p.push_back(Point(x,y+w)); p.push_back(Point(x+w,y+w)); } int ans=diameter(p); //int ians=floor(ans+0.5); printf("%d\n",ans); } return 0;} 0 0
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