LA 3263 - That Nice Euler Circles（直线相交以及交点）

简介：一笔画分割平面

分析：

我们要是直接计算平面的个数就会很难受

这时候我们注意到了题目中有Euler这个字眼

这就提示我们要使用欧拉定理

欧拉定理：

设平面图的顶点数，边数和面数分别为V,E,F，则V+F-E=2

这样我们只要求出顶点数和边数，就可以得到面数了：F=E-V+2

该平面的顶点由两部分组成：原先给出的点和直线之间的交点

注意，由于可能出现三点共线，所以需要删除多余的点

tip

使用unique的时候，需要重载 < 和 ==

node operator + (const node &a,const node &b){return node(a.x+b.x,a.y+b.y);}
node operator - (const node &a,const node &b){return node(a.x-b.x,a.y-b.y);}
node operator * (const node &a,const double &b){return node(a.x*b,a.y*b);}
node operator / (const node &a,const double &b){return node(a.x/b,a.y/b);}
bool operator < (const node &a,const node &b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator == (const node &a,const node &b){return a.x==b.x&&a.y==b.y;}

一坨重载，估计这是比较全的了

//这里写代码片
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>

using namespace std;

const double eps=1e-8;
const int N=310;
struct node{
double x,y;
node (double xx=0,double yy=0)
{
x=xx;y=yy;
}
};
node po
,V
;

node operator + (const node &a,const node &b){return node(a.x+b.x,a.y+b.y);}
node operator - (const node &a,const node &b){return node(a.x-b.x,a.y-b.y);}
node operator * (const node &a,const double &b){return node(a.x*b,a.y*b);}
node operator / (const node &a,const double &b){return node(a.x/b,a.y/b);}
bool operator < (const node &a,const node &b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
bool operator == (const node &a,const node &b){return a.x==b.x&&a.y==b.y;}

int dcmp(double x)
{
if (fabs(x)<eps) return 0;
else if (x>0) return 1;
else return -1;
}

double Cross(node x,node y){return x.x*y.y-x.y*y.x;}
double Dot(node x,node y){return x.x*y.x+x.y*y.y;}

node JiaoDian(node P,node v,node Q,node w)
{
node u=P-Q;
double t=Cross(w,u)/Cross(v,w);
return P+v*t;
}

bool XiangJiao(node a1,node a2,node b1,node b2)
{
double c1=Cross(a2-a1,b1-a1);
double c2=Cross(a2-a1,b2-a1);
double c3=Cross(b2-b1,a1-b1);
double c4=Cross(b2-b1,a2-b1);
return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}

bool Onit(node p,node a1,node a2)
{
return dcmp(Cross(a1-p,a2-p))==0 && dcmp(Dot(a1-p,a2-p))<0;
}

int main()
{
int n;
int cnt=0;
while (scanf("%d",&n)!=EOF&&n)
{
for (int i=0;i<n;i++)
{
scanf("%lf%lf",&po[i].x,&po[i].y);
V[i]=po[i];
}

n--;
int c=n,e=n;     // 初始点数和边数

for (int i=0;i<n;i++)
for (int j=i+1;j<n;j++)
if (XiangJiao(po[i],po[i+1],po[j],po[j+1]))       //线段相交
V[++c]=JiaoDian(po[i],po[i+1]-po[i],po[j],po[j+1]-po[j]);

sort(V,V+c);
c=unique(V,V+c)-V;     //去重

for (int i=0;i<c;i++)
for (int j=0;j<n;j++)
if (Onit(V[i],po[j],po[j+1])) e++;

printf("Case %d: There are %d pieces.\n",++cnt,e-c+2);
}
return 0;
}