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 = 300 + 10;

struct Point
{
    double x, y;
    Point(double x=0, double y=0) :x(x),y(y) {}
};
typedef Point Vector;
const double EPS = 1e-10;

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 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 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)) < 0)    return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0)    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)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool OnSegment(Point P, Point a1, Point a2)
{
    Vector v1 = a1 - P, v2 = a2 - P;
    return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0;
}

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

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

    int n, kase = 0;
    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 0; 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 = 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.
", ++kase, e+2-c);
    }

    return 0;
}