UVALive 3263 That Nice Euler Circuit

   给出一个点的序列，按顺序做一笔画，求最后的图形把平面分成了几部分。

   欧拉定理：E：线段数，V节点数，F平面数；有F+V-E==2.

   所以存下来所有线段之后，求出不同的交点数，再根据这些交点找出分割后的线段数，最后又公式算出F即可。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <string>typedef double type;using namespace std;struct Point{    type x,y;    Point(){}    Point(type a,type b)    {        x=a;        y=b;    }    void read()    {        scanf("%lf%lf",&x,&y);    }    void print()    {        printf("%.6lf %.6lf",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,type p){    return Vector(A.x*p,A.y*p);}Vector operator / (Vector A,type 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);}const double eps=1e-10;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;}//atan2(x,y) :向量(x,y)的极角，即从x轴正半轴旋转到该向量方向所需要的角度。type Dot(Vector A,Vector B){    return A.x*B.x+A.y*B.y;}type Cross(Vector A,Vector B){    return A.x*B.y-A.y*B.x;}type Length(Vector A){    return sqrt(Dot(A,A));}type Angle(Vector A,Vector B){    return acos(Dot(A,B))/Length(A)/Length(B);}type 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)//单位法线,左转90度，长度归一{    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=B-A,v2=P-A;    return fabs(Cross(v1,v2))/Length(v1);}double DistanceToSegment(Point P,Point A,Point B){    if (A==B) return Length(P-A);    Vector 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);}Point GetLineProjection(Point P,Point A,Point B)//P在AB上的投影{    Vector v=B-A;    return A+v*(Dot(v,P-A)/Dot(v,v));}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;}double ConvexPolygonArea(Point* p,int n)//多边形面积{    double area=0;    for (int i=1; i<n-1; i++)    area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2.0;}double PolygonArea(Point* p,int n)//有向面积{    double area=0;    for (int i=1; i<n-1; i++)    area+=Cross(p[i]-p[0],p[i+1]-p[0]);    return area/2.0;}Point p[1020];Point ans[123000];int cnt;int n;int main(){//    freopen("in.txt","r",stdin);    int tt=0;    while(~scanf("%d",&n) && n)    {        tt++;        for (int i=0; i<n; i++)        p[i].read();        n--;        cnt=n;        memcpy(ans,p,sizeof p);        for (int i=0; i<n; i++)         for (int j=i+1; j<n; j++)         {             if (SegmentProperIntersection(p[i],p[i+1],p[j],p[j+1]))             {                 ans[cnt++]=GetLineIntersection(p[i],p[i+1]-p[i],p[j],p[j+1]-p[j]);             }         }        sort(ans,ans+cnt);        cnt=unique(ans,ans+cnt)-ans;        int e=n;        for (int i=0; i<cnt; i++)         for (int j=0; j<n; j++)         {             if (OnSegment(ans[i],p[j],p[j+1]))             {                e++;             }         }        printf("Case %d: There are %d pieces.\n",tt,e+2-cnt);    }    return 0;}