题意
分析
欧拉定理:设平面内顶点数、边数、面数分别为\(V,E,F\),则\(V+F-E=2\)。
枚举每对线段求交点,注意去重。
另外注意第n个端点和第一个端点重合。
时间复杂度\(o(T n^3)\)。
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<ctime>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
rg T data=0;
rg int w=1;
rg char ch=getchar();
while(!isdigit(ch))
{
if(ch=='-')
w=-1;
ch=getchar();
}
while(isdigit(ch))
{
data=data*10+ch-'0';
ch=getchar();
}
return data*w;
}
template<class T>T read(T&x)
{
return x=read<T>();
}
using namespace std;
typedef long long ll;
co 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<(co Point&rhs)co
{
return x<rhs.x||(x==rhs.x&&y<rhs.y);
}
bool operator==(co Point&rhs)co
{
return dcmp(x-rhs.x)==0&&dcmp(y-rhs.y)==0;
}
};
typedef Point Vector;
Vector operator+(Vector A,Vector B)
{
return Vector(A.x+B.x,A.y+B.y);
}
Vector operator-(Point A,Point B)
{
return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(Vector A,double p)
{
return Vector(A.x*p,A.y*p);
}
Vector operator/(Vector A,double p)
{
return Vector(A.x/p,A.y/p);
}
double Dot(Vector A,Vector B)
{
return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
return sqrt(Dot(A,A));
}
double Angle(Vector A,Vector B)
{
return acos(Dot(A,B)/Length(A)/Length(B));
}
double Cross(Vector A,Vector B)
{
return A.x*B.y-A.y*B.x;
}
double Area2(Point A,Point B,Point C)
{
return Cross(B-A,C-A);
}
Vector Rotate(Vector A,double rad)
{
return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A)
{
double L=Length(A);
return Vector(-A.y/L,A.x/L);
}
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;
}
double DistanceToLine(Point P,Point A,Point B)
{
Vector v1=B-A,v2=P-A;
return fabs(Cross(v1,v2))/Length(v1);
}
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);
if(dcmp(Dot(v1,v3))>0)
return Length(v3);
return DistanceToLine(P,A,B);
}
Point GetLineProjection(Point P,Point A,Point B)
{
Vector v=B-A;
return A+v*(Dot(v,P-A)/Dot(v,v));
}
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;
}
bool OnSegment(Point p,Point a1,Point a2)
{
return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dot(a1-p,a2-p))<0;
}
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;
}
co int N=300;
Point P[N],V[N*N];
int main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n,kase=0;
while(read(n))
{
for(int i=0;i<n;++i)
{
read(P[i].x);read(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;
}