That Nice Euler Circuit UVALive - 3263

题意：平面上有n个端点的一笔画，求这些线段将平面分成几个部分 。
分析：直接使用欧拉定理：V+F-E=2
V：结点数（两个边相交就算一个结点，要注意去重，加上新增的结点，去掉重复的点）
F：面数
E：边数（一个线段算是一条边）#include <bits/stdc++.h>

using namespace std;

struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y) {}
};

typedef Point Vector ;
const double esp=1e-10;
const int maxn=350;
Point p[maxn],v[maxn*maxn];
///两个向量相加
Vector operator +(Vector v1,Vector v2)
{
return Vector(v1.x+v2.x,v1.y+v2.y);
}
///两个向量相减
Vector operator -(Vector v1,Vector v2)
{
return Vector(v1.x-v2.x,v1.y-v2.y);
}
///向量与数相乘
Vector operator *(Vector v,double a)
{
return Vector(v.x*a,v.y*a);
}
Vector operator /(Vector v,double a)
{
return Vector(v.x/a,v.y/a);
}
///向量与数相除
bool operator<(Vector v1,Vector v2)
{
return v1.x<v2.x||(v1.x==v2.x&&v1.y<v2.y);
}
///减少精度问题
int dcmp(double x)
{
if(fabs(x)<esp) return 0;
return x<0?-1:1;
}
///判断两个向量是否相等
bool operator ==(Vector v1,Vector v2)
{
return dcmp(v1.x-v2.x)==0&&dcmp(v1.y-v2.y)==0;
}
///点积
double Dot(Vector v1,Vector v2)
{
return v1.x*v2.x+v1.y*v2.y;
}
double Length(Vector v)
{
return sqrt(Dot(v,v));
}

///叉积
double Cross(Vector v1,Vector v2)
{
return v1.x*v2.y-v1.y*v2.x;
}

///直线相交,求两条直线的交点
///调用前需保证两直线有唯一的交点，且不重合
Vector 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;
}

///判断线段相交
///充要条件：每条线段的端点都在另一条线段的两侧（叉积符号不同）
bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
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;
}
///判断点是否在线段上
bool OnSegment(Point p,Point a1,Point a2)
{
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
int n;
bool cmp(Point v1,Point v2)
{
return v1.x<v2.x||(v1.x==v2.x&&v1.y<v2.y);
}
int main()
{
int ka=0;
while(cin>>n)
{
if(n==0) break;
ka++;

for(int i=0; i<n; i++)
{
cin>>p[i].x>>p[i].y;
v[i]=p[i];
}
int c=n-1;
for(int i=0; i<n-1; i++)
{
for(int j=i+1; j<n-1; 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,cmp);
c=unique(v,v+c)-v;
int e=n-1;
for(int i=0; i<c; i++)
{
for(int j=0; j<n-1; j++)
{
if(OnSegment(v[i],p[j],p[j+1])) e++;
}
}
cout<<"Case "<<ka<<": There are "<<e+2-c<<" pieces."<<endl;
}
}