Uvalive 3263 That Nice Euler Circuit（几何欧拉定理）

题目地址：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1264

思路：欧拉定理：设平面图的顶点数、边数和面数分别为V、E和F，则V+F-E=2。

顶点数可以通过计算两线段交点计算：首先判断两线段是否相交（每条线段的两个端点均在另一条线段的两侧），若相交，则按照求两直线交点方式求出两线段交点。由于可能有多于两条线段交于一点，所以交点需判重。

边数可以根据所求顶点，枚举每条边，判断顶点是否位于该条边上（非端点位置）。若有一点位于边上，则每次将该条线段分隔为两部分，边数加一。

#include<cmath>#include<cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#define debugusing namespace std;const double eps=1e-10;struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) {}    void read()    {        scanf("%lf%lf",&x,&y);    }    void print()    {      printf("%.6f %.6f",x,y);    }};typedef Point Vector;Vector operator + (Vector A,Vector B){    return Vector(A.x+B.x,A.y+B.y);}Vector operator - (Point A,Point 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);}bool operator < (const Point &a,const Point &b){    return a.x<b.x||(a.x==b.x&&a.y<b.y);}int dcmp(double x){    if(fabs(x)<eps) return 0;    else return x<0?-1:1;}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 Length(Vector A){    return sqrt(Dot(A,A));}double Angle(Vector A,Vector B){    return acos(Dot(A,B)/Length(A)/Length(B));}Vector Rotate(Vector A,double rad){    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}double Cross(Vector A,Vector B){    return A.x*B.y-A.y*B.x;}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;}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point 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;}bool OnSegment(Point p,Point a1,Point a2){  return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;}const int maxn=300+50;int n;Point p[maxn];vector<Point> totp;int main(){#ifdef debu    freopen("in.txt","r",stdin);#endif // debug    int cas=0;    while(scanf("%d",&n)!=EOF&&n)    {      int numl=n-1;      totp.clear();      for(int i=1;i<=n;i++)      {        p[i].read();        totp.push_back(p[i]);      }      for(int i=1;i<n;i++)      {        for(int j=i+1;j<n;j++)        {          if(SegmentProperIntersection(p[i],p[i+1],p[j],p[j+1]))          {            totp.push_back(GetLineIntersection(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]));          }        }      }      sort(totp.begin(),totp.end());      int nump=unique(totp.begin(),totp.end())-totp.begin();      for(int i=0;i<nump;i++)      {        for(int j=1;j<n;j++)        {          if(OnSegment(totp[i],p[j],p[j+1]))   //判断点是否在一条线段上（不含端点）           {             numl++;           }        }      }      printf("Case %d: There are %d pieces.\n",++cas,numl-nump+2);    }    return 0;}