That Nice Euler Circuit
【题目链接】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.
",++kase,e+2-c);
}
return 0;
}