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，因此团史一条闭合曲线。组成一笔画的线段可以相交，但是不会部分重叠。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）。

分析：若是直接找出所有区域，或非常麻烦，而且容易出错。但用欧拉定理可以将问题进行转化，使解法变容易。

欧拉定理：设平面图的顶点数、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。

设平面图的结点由两部分组成，即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。

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 using namespace std;
 const int inf=0x3f3f3f3f;
 const double eps = 1e-;
 const int N =  + ;

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

 int dcmp(double x)
 {
     if(fabs(x) < eps)
         return ;
     else
     return x <  ? - : ;
 }

 bool operator == (const Point& a, const Point& b)
 {
     return dcmp(a.x - b.x) ==  && dcmp(a.y - b.y) == ;
 }

 double Dot(Vector A, Vector B)
 {
     return A.x * B.x + A.y * B.y;
 }

 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) <  && dcmp(c3) * dcmp(c4) < ;
 }

 bool OnSegment(Point p, Point a1, Point a2)
 {
     return dcmp(Cross(a1-p, a2-p)) ==  && dcmp(Dot(a1-p, a2-p)) < ;
 }

 Point P[N], V[N*N];

 int main()
 {
     int n, cas = ;
     while(~scanf("%d",&n) && n)
     {
         for(int i = ; i < n; i++)
         {
             scanf("%lf%lf", &P[i].x, &P[i].y);
             V[i] = P[i];
         }
         n--;
         int vcnt = n, ecnt = n;
         for(int i = ; i < n; i++)
             for(int j = i + ; j < n; j++)
             {
                 if(SegmentProperIntersection(P[i], P[i+], P[j], P[j+]))
                     V[vcnt++] = GetLineIntersection(P[i], P[i+]-P[i], P[j], P[j+]-P[j]);
             }
         sort(V, V+vcnt);
         vcnt = unique(V, V+vcnt) - V;//去掉相邻元素中重复的，使用前先排序
         for(int i = ; i < vcnt; i++)
             for(int j = ; j < n; j++)
                 if(OnSegment(V[i], P[j], P[j+]))
                     ecnt++;
         int ans = ecnt +  - vcnt;
         printf("Case %d: There are %d pieces.\n", ++cas, ans);
     }
     return ;
 }