UVa1342 That Nice Euler Circuit (计算几何)

题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=446&page=show_problem&problem=4088

使用平面图的欧拉定理，(V+F-E=2)，所以问题就变成求出有多少个顶点和边

枚举线段，判段线段两两是否产生新的交点（端点相交不算），可能出现三线共点的情况，所以还需要去重

原来的边上每新增一个点，则边数就加一，判断新增点是否在原来的线段上即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 310; 

const double eps = 1e-10;
int dcmp(double x) {
  if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

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

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

bool operator == (const Point& a, const Point &b) {
  return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

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); }
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;
}

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));
}

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;
}

int n, tot, e;
Point p[maxn], v[maxn*maxn];

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int kase = 0;
	while(scanf("%d", &n) == 1 && n){
		tot = 0;
		for(int i = 1 ; i <= n ; ++i){
			scanf("%lf%lf", &p[i].x, &p[i].y);
			v[i] = p[i];
		}
		--n;
		tot = n; e = n;
		
		for(int i = 1 ; i <= n ; ++i){
			for(int j = i+1 ; j <= n ; ++j){
				if(SegmentProperIntersection(p[i], p[i+1], p[j], p[j+1])){
					v[++tot] = GetLineIntersection(p[i], p[i+1]-p[i], p[j], p[j+1]-p[j]);
				}
			}
		}
		
		sort(v+1, v+1+tot);
		tot = unique(v+1, v+1+tot) - v - 1;
		
		for(int i = 1 ; i <= tot ; ++i){
			for(int j = 1 ; j <= n ; ++j){
				if(OnSegment(v[i], p[j], p[j+1])) ++e;
			}
		}
		
		printf("Case %d: There are %d pieces.
", ++kase, e+2-tot);
	}
	return 0;
}