UVALive 3263 That Nice Euler Circuit

题意：平面上有一个包含n个端点的一笔画，第n个端点总是和第一个端点重合，因此图案是一条闭合曲线。组成一笔画的线段可以相交，但是不会部分重叠，给出各个顶点的坐标，求这些线段将平面分成多少部分（包括封闭区域和无限大区域）

解题思路：用欧拉定理将问题进行转化。欧拉定理：设平面内的顶点数、边数和面数分别为V，E，F，则V+F-E=2。这样，只需求出顶点数V和边数E，就可以求出F=E+2-V。该平面图的结点有两部分组成，及原来的结点和新增的结点。由于可能出现三线共点，需要删除重复的点

代码：

#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;

struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(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);
}
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;
}
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));
}
double Cross(Vector A,Vector B)
{
return A.x*B.y-A.y*B.x;
}
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));
}
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=310;
Point P[maxn],V[maxn*maxn];
int main()
{
int n;
int cas=0;
while(scanf("%d",&n)&&n)
{
for(int i=0; i<n; i++)
{
scanf("%lf%lf",&P[i].x,&P[i].y);
V[i]=P[i];
}
int c=n,e=n-1;//原来的顶点数和边数
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]))//两线段相交
{
V[c++]=GetLineIntersection(P[i],P[i+1]-P[i],P[j],P[j+1]-P[j]);//交点
}
}
}
sort(V,V+c);
c=unique(V,V+c)-V;//去重点
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++;//点在边上，边数就增加了
}
}
printf("Case %d: There are %d pieces.\n",++cas,e+2-c);
}
return 0;
}