UVa 1342 That Nice Euler Circuit(几何)

题目链接：UVa 1342 That Nice Euler Circuit

几何。

欧拉定理，设平面图的顶点数、边数、和面数分别为V，E和F，则V+F-E=2。

然后求出来顶点数和边数就可以了。

主要还是用了白书上的几何模版，很好很强大。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>

using namespace std;

const double eps = 1e-10;
const int MAX_N = 300 + 10;

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

typedef Point Vector;

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

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

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

double Length(const Vector& A)
{
    return sqrt(Dot(A, A));
}

double Angle(const Vector& A, const Vector& B)
{
    return acos(Dot(A, B) / Length(A) / Length(B));
}

double Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const 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(const Point& p, const Point& a1, const Point& a2)
{
	return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}
Point GetLineIntersection(const Point& P, const Point& v, const Point& Q, const Point& w)
{
	Vector u = P-Q;
	double t = Cross(w, u) / Cross(v, w);
	return P+v*t;
}

Vector Rotate(const Vector& A, double rad)
{
	return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}


Point p[MAX_N], v[MAX_N * MAX_N];
int n, cnt;

int main()
{
    cnt = 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];
        }
        n--;
        int c, e;
        c = e = n;
        //枚举边添加点
        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]);
            }
        }
        //可能出现三线共点，所以需要删除重复的点
        sort(v, v + c);
        c = unique(v, v + c) - v;//返回不重复元素的数目
        /**添加边
        添加新边，OnSegment判断点v[i]是否在p[j]和p[j + 1]组成的线段上，不包括端点。
        枚举的边都是原有边，
        那么如果一个点在线段上，这个点就把这条线段分成了两段，所以边数需要加1，
        如果这个点是这条边的端点，边数不需要加1，其实就是说明这个点是原有点，而原有点不会分割原有线段
        这个逻辑也正符合OnSegment的判定规则。
        **/
        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.\n", ++cnt, e - c + 2);
    }
    return 0;
}