LA 4728 Squares 旋转卡壳
2014-04-19 07:16
344 查看
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
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; }
相关文章推荐
- LA 4728 Squares (二维凸包+旋转卡壳)
- LA 4728 (旋转卡壳) Squares
- LA 4728凸包算法-旋转卡壳的直径
- Uvalive 4728 Squares(旋转卡壳)
- UVALive 4728 Squares(旋转卡壳求凸包直径)
- 【LA 4728】Square, Seoul 2009 (凸包,旋转卡壳)
- LA 4728 旋转卡壳算法求凸包的最大直径
- 旋转卡壳,凸包直径(正方形,LA 4728)
- UVALive 4728 Squares(旋转卡壳)
- LA 4728 Square ,旋转卡壳法求多边形的直径
- UVA 4728 Squares(凸包+旋转卡壳)
- LA 4728 Squares 凸包 .
- uva 1453 - Squares(旋转卡壳)
- uva 1453 Squares (旋转卡壳 rotating calipers)
- 凸包扫描 + 旋转卡壳 UVALive 4728
- UVa 1453 - Squares 旋转卡壳求凸包直径
- uvalive 4728(旋转卡壳求凸包最长直径)
- UVALive 4728 (凸包 旋转卡壳)
- UVA LA 4728旋转卡壳
- LA 1453 旋转卡壳