LA 3263 (平面图的欧拉定理) That Nice Euler Circuit

题意：

平面上有n个端点的一笔画，最后一个端点与第一个端点重合，即所给图案是闭合曲线。求这些线段将平面分成多少部分。

分析：

平面图中欧拉定理：设平面的顶点数、边数和面数分别为V、E和F。则 V+F-E=2

所求结果不容易直接求出，因此我们可以转换成 F=E-V+2

枚举两条边，如果有交点则顶点数+1，并将交点记录下来

所有交点去重(去重前记得排序)，如果某个交点在线段上，则边数+1

 //#define LOCAL
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;

 const int maxn =  + ;

 struct Point
 {
     double x, y;
     Point(double x=, double y=) :x(x),y(y) {}
 };
 typedef Point Vector;
 const double EPS = 1e-;

 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }

 Vector operator - (Vector A, Vector 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 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 VRotate(Vector A, double rad)
 {
     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
 }

 Point PRotate(Point A, Point B, double rad)
 {
     return A + VRotate(B-A, rad);
 }

 Vector Normal(Vector A)
 {
     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)) < )    return Length(v2);
     else if(dcmp(Dot(v1, v3)) > )    return Length(v3);
     else return fabs(Cross(v1, v2)) / Length(v1);
 }

 Point GetLineProjection(Point P, Point A, Point B)
 {
     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);
     double 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)
 {
     Vector v1 = a1 - P, v2 = a2 - P;
     return dcmp(Cross(v1, v2)) ==  && dcmp(Dot(v1, v2)) < ;
 }

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

 int main(void)
 {
     #ifdef    LOCAL
         freopen("3263in.txt", "r", stdin);
     #endif

     int n, kase = ;
     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 c = n, e = 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[c++] = GetLineIntersection(P[i], P[i+]-P[i], P[j], P[j+]-P[j]);

         sort(V, V+c);
         c = unique(V, V+c) - V;

         for(int i = ; i < c; ++i)
             for(int j = ; j < n; ++j)
                 if(OnSegment(V[i], P[j], P[j+]))    e++;

         printf("Case %d: There are %d pieces.\n", ++kase, e+-c);
     }

     return ;
 }

代码君