That Nice Euler Circuit

【题目类型】几何

&题解：

蓝书P260 要用欧拉定理:V+F=E+2 V是顶点数;F是分成了多少区域,也就是本题的答案;E是有多少条边,比如2条线段相交,就有4条边,而不是2条.

还有几点注意:

1.dcmp()没有返回0 调了半天(模板照着敲都能错 0.0!)

2.V[]点没有去重 wa了1次(这个去重还是很难想的,去重之后还要证出原来的方法是正确的)

还有他这种算E(边数)的想法很好:就是枚举最开始给的那n条边,在每条边上找有几个点,假设有1个点在这条边上,这1条边就变成了2条边,以此类推.所以前面就要把已知的点和产生的交点找到,最后别忘了去重.

&代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

//蓝书P255

//1.点的定义

struct Point {

	double x,y;

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

};

//点和向量是一样的内容 所以会出来2个名字

typedef Point Vector;

//向量+向量=向量, 点+向量=点

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

//点-点=向量

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

//向量*数=向量

Vector operator * (Vector A,double p) {return Vector(A.x*p,A.y*p);}

//向量/数=向量

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);}

const double eps=1e-10;

//doublue的三态函数

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;}

//点积

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

//向量长度

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

//向量夹角

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

//叉积

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

//有向面积,数值是三角形面积的2倍

double Area2(Point A,Point B,Point C) {return Cross(B-A,C-A);}

//向量旋转,doublue的都是弧度

Vector Rotate(Vector A,double rad) {

	return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));

}

//求单位法向量

Vector Normal(Vector A) {

	double L=Length(A);

	return Vector(-A.y/L,A.x/L);

}

//2.点和直线

//调用前请确保两条直线P+tv和Q+tw有唯一交点.当且仅当Cross(v,w)非0

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;

}

//点P到直线(过点A,B)的距离

double DistanceToLine(Point P,Point A,Point B) {

	Vector v1=B-A,v2=P-A;

	return fabs(Cross(v1,v2))/Length(v1);//不加fabs,得到的是有向距离

}

//点P到线段AB的距离

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);

}

//求点P在直线AB的投影点Q

Point GetLineProjection(Point P,Point A,Point B) {

	Vector v=B-A;

	return A+v*(Dot(v,P-A)/Dot(v,v));

}

//线段相交判定,两线段恰好有一个公共点,且不在任何一条线段的端点

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)<0&&dcmp(c3)*dcmp(c4)<0;

}

//判断点p是否在线段a1a2上(不包含线段的端点)

bool OnSegment(Point p,Point a1,Point a2) {

	return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;

	// return dcmp(Dot(a1-p,a2-p))<0;

}

//多边形的有向面积

double PolygonArea(Point* p,int n) {

	double area=0;

	for(int i=1; i<n-1; i++) {

		area+=Cross(p[i]-p[0],p[i+1]-p[0]);

	}

	return area/2;

}

//解题代码

const int maxn= 300 +7;

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

//dcmp()没有返回0 调了半天

//V[]点没有去重 wa了1次

int main() {

	freopen("E:1.in","r",stdin);

	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.\n",++kase,e+2-c);

	}

	return 0;

}

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UVALi 3263 That Nice Euler Circuit(几何)

That Nice Euler Circuit