求多邊形的核，直接把所有邊求半平面交判斷有無即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=205;
const double eps=1e-6;
int n;
struct dian
{
    double x,y;
    dian(double X=0,double Y=0)
    {
        x=X,y=Y;
    }
    dian operator + (const
 
 dian &a)
    {
        return dian(x+a.x,y+a.y);
    }
    dian operator - (const dian &a)
    {
        return dian(x-a.x,y-a.y);
    }
    dian operator * (const double &a) const
    {
        return dian(x*a,y*a);
    }
    dian operator / (const double &a) const
    {
        return
 
 dian(x/a,y/a);
    }
}p[N];
struct bian
{
    dian s,v;
    bian(dian S=dian(),dian V=dian())
    {
        s=S,v=V;
    }
}l[N],s[N];
int read()
{
    int r=0,f=1;
    char p=getchar();
    while(p>‘9‘||p<‘0‘)
    {
        if(p==‘-‘)
            f=-1;
        p=getchar();
    }
    while(p>=‘0‘
 
&&p<=‘9‘)
    {
        r=r*10+p-48;
        p=getchar();
    }
    return r*f;
}
double cj(dian a,dian b)
{
    return a.x*b.y-a.y*b.x;
}
double mj(dian a,dian b,dian c)
{
    return cj(b-a,c-a)/2;
}
dian jd(bian x,bian y)
{
    return x.s+x.v*(cj(x.s-y.s,y.v)/cj(y.v,x.v));
}
bool px(bian a,bian b)
{
    return cj(a.v,b.v)==0;
}
bool bn(bian a,bian b)
{
    int ar=cj(a.v,b.v);
    return ar>0||(ar==0&&cj(a.v,b.s-a.s)>0);
}
bool dn(dian x,bian y)
{
    return cj(y.v,x-y.s)<=0;
}
bool cmp(const bian &x,const bian &y)
{
    if(x.v.y==0&&y.v.y==0)
        return x.v.x<y.v.x;
    if((x.v.y<=0)==(y.v.y<=0))
        return bn(x,y);
    return x.v.y<y.v.y;
}
double dis2(bian a)
{
    return sqrt(a.v.x*a.v.x+a.v.y*a.v.y);
}
bian yi(bian a,double r)
{
    return bian(dian(a.s.x-a.v.y*r/dis2(a),a.s.y+a.v.x*r/dis2(a)),a.v);
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
            p[i].x=read(),p[i].y=read();
        p[n+1]=p[1];
        for(int i=1;i<=n;i++)
            l[i]=bian(p[i],p[i+1]-p[i]);sort(l+1,l+1+n,cmp);
        int top=0;
        for(int i=1;i<=n;i++)
            if(i==1||!px(l[i],l[i-1]))
                l[++top]=l[i];
        n=top;
        int ll=1,rr=2;
        s[1]=l[1],s[2]=l[2];
        for(int i=3;i<=n;i++)
        {
            while(ll<rr&&dn(jd(s[rr],s[rr-1]),l[i]))
                rr--;
            while(ll<rr&&dn(jd(s[ll],s[ll+1]),l[i]))
                ll++;
            s[++rr]=l[i];
        }
        while(ll<rr&&dn(jd(s[rr],s[rr-1]),s[ll]))
            rr--;
        printf("%d\n",(rr-ll>1));
    }
    return 0;
}

poj 3130 How I Mathematician Wonder What You Are! 【半平面交】