const double eps = 1e-8;
const double PI = acos(-1.0);

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

struct Point
{
    double x,y;
    Point() {}
    Point(double a,double b):x(a),y(b) {}
    Point operator -(const Point &b)const
    {
        return Point(x-b.x, y-b.y);
    }
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};

struct Line
{
    Point s, e;
    double k;
    Line() {}
    Line(Point a, Point b)
    {
        s = a;
        e = b;
        k = atan2(e.y - s.y,e.x - s.x);
    }
    Point operator &(const Line &b)const //获得两个向量交点
    {
        Point res = s;
        double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
        res.x += (e.x-s.x)*t;
        res.y += (e.y-s.y)*t;
        return res;
    }
} Q[110]; //队列

//半平面交，直线的左边代表有效区域
bool HPIcmp(Line a, Line b)//极角排序
{
    if(fabs(a.k - b.k) > eps)
        return a.k < b.k;//如果相同保留最左侧的一个
    return ((a.s - b.s)^(b.e - b.s)) < 0;
}

void HPI(Line line[], int n, Point res[], int &resn)
{
    int tot = n;
    sort(line, line+n, HPIcmp);
    tot = 1;
    for(int i = 1; i < n; i++) //去重
        if(fabs(line[i].k - line[i-1].k) > eps)
            line[tot++] = line[i];

    int head = 0, tail = 1;
    Q[0] = line[0];
    Q[1] = line[1];
    resn = 0;
    for(int i = 2; i < tot; i++)
    {
        if(fabs((Q[tail].e-Q[tail].s)^(Q[tail-1].e-Q[tail-1].s)) < eps 
        || fabs((Q[head].e-Q[head].s)^(Q[head+1].e-Q[head+1].s)) < eps)    return;
        while(head < tail && (((Q[tail]&Q[tail-1]) - line[i].s)^(line[i].e-line[i].s)) > eps)    tail--;
        while(head < tail && (((Q[head]&Q[head+1]) - line[i].s)^(line[i].e-line[i].s)) > eps)    head++;
        Q[++tail] = line[i];
    }
    while(head < tail && (((Q[tail]&Q[tail-1]) - Q[head].s)^(Q[head].e-Q[head].s)) > eps)   tail--;
    while(head < tail && (((Q[head]&Q[head-1]) - Q[tail].s)^(Q[tail].e-Q[tail].e)) > eps)   head++;
    if(tail <= head + 1)return;
    for(int i = head; i < tail; i++)   res[resn++] = Q[i]&Q[i+1];
    if(head < tail - 1)   res[resn++] = Q[head]&Q[tail];

    ///计算核心面积
    double ares = 0;
    for (int i = 0; i < resn; i++)
        ares += Cross(res[i],res[(i+1)%resn]) / 2;
    ares = fabs(ares);
}