That Nice Euler Circuit UVA - 1342

思路：欧拉定理：平面的  顶点数+面数-边数=2

该平面节点分为两部分，即原点的结点和新增的结点。由于更能出线三点共线，需要删除重复的点。 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<cassert>

using namespace std;
struct Point {
    double x, y;
    Point (double x=0, double y=0):x(x), y(y){}
};
bool operator <(const Point&a,const Point&b){//从小到大排序
    return a.x<b.x || (a.x==b.x&&a.y<b.y);
}
typedef Point Vector;
//重载向量间的加减乘除
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); }

Point read_point(){//读入点
    Point p;
    scanf("%lf%lf", &p.x, &p.y);
    return p;
}

double Dot(Vector A, Vector B){return A.x*B.x+A.y*B.y; }//A，B点乘
double Length(Vector A){ return sqrt(Dot(A, A)); }//求向量长度

double Angle(Vector A, Vector B){//求两向量夹角
    return acos(Dot(A, B)/Length(A)/Length(B));
}

Vector Rotate(Vector A, double rad){//向量A逆时针转rad后的向量值
    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}

double Cross(Vector v1, Vector v2){//叉乘
    return v1.x*v2.y-v1.y*v2.x;
}
Point GetLineIntersection(Point A, Vector v1, Point B, Vector v2){//求在点v1上的向量A和在点v2上的向量B的交点
    Vector v3=A-B;
    double t=Cross(v2, v3)/Cross(v1, v2);
    return A+v1*t;
}
double DistanceToLine(Point P,Point A,Point B){//点P到线A，B的距离
    Vector v1=B-A;
    Vector v2=P-A;
    return fabs(Cross(v1,v2)/Length(v1));
}
Point GetLineProjection(Point P,Point A,Point B){//点P到线A，B的投影的点
    Vector v=B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}
const double eps=1e-10;//判断浮点数和0的关系
int dcmp(double x){
    if(fabs(x)<eps)return 0;
    else return x<0?-1:1;
}
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){//判断线a1,b1和线a2,b2是否相交或判断两点是否同侧
    double c1=Cross(a2-a1,b1-a1),c2=Cross(a2-a1,b2-a1);
    double c3=Cross(b2-b1,a1-b1),c4=Cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
}
double ConvexArea(Point *p,int n){//tu多边形有向面积
    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;
}
bool Onsegment(Point p,Point a1,Point a2){//判断三个点是否在一条直线上
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
bool operator ==(const Point&a,const Point&b){//判断是否相等，去重的时候用
    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
const int maxn=300+10;
Point p[maxn],V[maxn*maxn];
int main(){
    int n,ca=0;
    while(scanf("%d",&n)&&n){
        for(int i=0;i<n;i++){
            p[i]=read_point();
            V[i]=p[i];
        }
        n--;
        long c=n,e=n;
        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])){
                    V[c++]=GetLineIntersection(p[i], p[i+1]-p[i], p[j], p[j+1]-p[j]);
                }
            }
        }
        sort(V, V+c);
        c=unique(V, V+c)-V;
        for(int i=0;i<c;i++){
            for(int j=0;j<n;j++){
                if(Onsegment(V[i], p[j], p[j+1]))e++;
            }
        }
        printf("Case %d: There are %d pieces.\n",++ca,e+2-c);
    }
    return 0;
}