poj 3449 Geometric Shapes 线段相交
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题目地址：poj3449
考点：输入正方形（边不平行于坐标轴）存储。
用string s[] 记录是什么形状， int vertex[] 记录顶点个数。char ch[] 记录字符标识是什么。
一个写起来比较繁琐的模拟题吧，思路没什么，主要是细心，注意输出trick
#include<iostream>#include<cmath>#include<cstdio>#include<algorithm>#include<vector>#include<cstring>#include<string>const double INF=0x3fffffff;const double eps=1e-10;const double PI=acos(-1.0);using namespace std;struct Point{ double x; double y; Point(double x=0,double y=0):x(x),y(y){} };int dcmp(double x) {return (x>eps)-(x<-eps); }int sgn(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,double p) { return Vector(A.x*p,A.y*p); }Vector operator /(Vector A,double p) {return Vector(A.x/p,A.y/p);}ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y; 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;}double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}double Cross(Vector A,Vector B) {return A.x*B.y-B.x*A.y; }double Length(Vector A) { return sqrt(Dot(A, A));}double Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}double Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Vector Normal(Vector A) {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; double t=Cross(w, u)/Cross(v,w); return P+v*t; }double DistanceToLine(Point P,Point A,Point B){ Vector v1=P-A; Vector v2=B-A; return fabs(Cross(v1,v2))/Length(v2); }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; double t=Dot(v,v1)/Dot(v,v); return A+v*t;}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){ double c1=Cross(b1-a1, a2-a1); double c2=Cross(b2-a1, a2-a1); double c3=Cross(a1-b1, b2-b1); 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;}bool SegmentCommon(Point a1,Point a2,Point b1,Point b2){ double c1=Cross(b1-a1, a2-a1); double c2=Cross(b2-a1, a2-a1); double c3=Cross(a1-b1, b2-b1); double c4=Cross(a2-b1, b2-b1); if(c1==0&&c2==0&&c3==0&&c4==0) { if(OnSegment(a1, b1, b2)) return 1; if(OnSegment(a2, b1, b2)) return 1; return 0; } return dcmp(c1)*dcmp(c2)<=0&&dcmp(c3)*dcmp(c4)<=0 ; }double PolygonArea(Point *p,int n){ 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; char ch; // scanf("%lf,%lf",&P.x,&P.y); cin>>ch>>P.x>>ch>>P.y>>ch; return P;}// ---------------与圆有关的--------struct Circle{ Point c; double r; Circle(Point c=Point(0,0),double r=0):c(c),r(r) {} Point point(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(double t) { return Point(p+v*t); } };int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol){ double a=L.v.x; double b=L.p.x-C.c.x; double c=L.v.y; double d=L.p.y-C.c.y; double e=a*a+c*c; double f=2*(a*b+c*d); double g=b*b+d*d-C.r*C.r; 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double angle(Vector v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol){ 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 double a=angle(C2.c-C1.c); 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; 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 { double 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 } double d=Length(A.c-B.c); double rdiff=A.r-B.r; double rsum=A.r+B.r; if(dcmp(d-rdiff)<0) return 0; // 内含 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 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 { 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;}// 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; }struct Segment{ Point A; Point B; Segment(Point a=Point(0,0),Point b=Point(0,0)) :A(a),B(b) {} };Point p[50][50];char ch[50];string s[50];int vertex[50];vector<char> v_ip[50];int main(){ char id; int i=0; while(cin>>id) { if(id=='.') break; for(int j=0;j<50;j++) v_ip[j].clear(); if(id=='-') { for(int k=0;k<i;k++) { for(int l=k+1;l<i;l++) { bool ok=0; for(int a=0;a<vertex[k];a++) for(int b=0;b<vertex[l];b++) { if(SegmentCommon(p[k][a], p[k][(a+1)%vertex[k]], p[l][b], p[l][(b+1)%vertex[l]])) { ok=1; break; } } if(ok) { v_ip[k].push_back(ch[l]); v_ip[l].push_back(ch[k]); } } } for(int j=0;j<i;j++) for(int k=j+1;k<i;k++) { if(ch[j]>ch[k]) { swap(ch[j],ch[k]); swap(v_ip[j],v_ip[k]); } } for(int j=0;j<i;j++) { cout<<ch[j]<<' '; if(v_ip[j].size()==0) { cout<<"has no intersections"<<endl; } else { sort(v_ip[j].begin(),v_ip[j].end()); cout<<"intersects with "; if(v_ip[j].size()==1) { cout<<v_ip[j][0]<<endl; } else if(v_ip[j].size()==2) { cout<<v_ip[j][0]<<" and "<<v_ip[j][1]<<endl; } else { for(int k=0;k<v_ip[j].size()-1;k++) { cout<<v_ip[j][k]<<", "; } cout<<"and "<<v_ip[j][v_ip[j].size()-1]<<endl; } } } cout<<endl; i=0;// 注意清零 } else { ch[i]=id; cin>>s[i]; if(s[i]=="square") { vertex[i]=4; Point A,B; A=read_point(); B=read_point(); p[i][0]=A; p[i][2]=B; Vector AB=B-A; Vector AB_nor=Rotate(AB, PI/2)/2; Point mid=(A+B)/2; p[i][1]=mid+AB_nor; p[i][3]=mid-AB_nor; } else if(s[i]=="rectangle") { vertex[i]=4; Point A,B,C; A=read_point(); B=read_point(); C=read_point(); p[i][0]=A; p[i][1]=B; p[i][2]=C; p[i][3]=(A+C)-B; } else if(s[i]=="line") { vertex[i]=2; p[i][0]=read_point(); p[i][1]=read_point(); } else if(s[i]=="triangle") { vertex[i]=3; p[i][0]=read_point(); p[i][1]=read_point(); p[i][2]=read_point(); } else if(s[i]=="polygon") { cin>>vertex[i]; for(int j=0;j<vertex[i];j++) { p[i][j]=read_point(); } } i++; } }} 0 0
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