半平面交（模板 详解后补）

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
#define N 50005

const double eps=1e-8;
int dcmp(double x) {
    if (x<=eps&&x>=-eps) return 0;
    return (x>0)?1:-1;
}
struct Vector {
    double x,y;
    Vector(double X=0,double Y=0){
        x=X,y=Y;
    }
};
typedef Vector Point;

struct Line {
    Point p;
    Vector v;
    double ang;
    Line(Point P=Point(0,0),Vector V=Vector(0,0)) {
        p=P,v=V;
        ang=atan2(v.y,v.x);
    }
    bool operator < (const Line &a) const {
        return ang<a.ang;
    }
};
Vector operator + (Vector a,Vector b) {return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b) {return Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,double b) {return Vector(a.x*b,a.y*b);}

int n,l,r,m,cnt;
double ans;
Line L[N],q[N];
Point p[N],poly[N];

double Cross(Vector a,Vector b) {
    return a.x*b.y-a.y*b.x;
}
Point GLI(Point P,Vector v,Point Q,Vector w) {
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
bool Onleft(Line m,Point P) {
    Vector w=P-m.p;
    return dcmp(Cross(m.v,w))>=0;
}
void halfp(){
    sort(L+1,L+n+1);
    cnt=0;
    q[l=r=1]=L[1];
    for (int i=2;i<=n;++i) {
        while (l<r&&!Onleft(L[i],p[r-1])) --r;
        while (l<r&&!Onleft(L[i],p[l])) ++l;
        q[++r]=L[i];
        if (dcmp(Cross(q[r].v,q[r-1].v))==0) {
            --r;
            if (Onleft(q[r],L[i].p))
                q[r]=L[i];
        }
        if (l<r)
            p[r-1]=GLI(q[r-1].p,q[r-1].v,q[r].p,q[r].v);
    }
    while (l<r&&!Onleft(q[l],p[r-1]))
        --r;
    if (r-l<=1) return;
    p[r]=GLI(q[r].p,q[r].v,q[l].p,q[l].v);
    for (int i=l;i<=r;++i) poly[++cnt]=p[i];
}
double Area() {
    double ans=0;
    for (int i=2;i<cnt;++i)
        ans+=Cross(poly[i]-poly[1],poly[i+1]-poly[1]);
    return fabs(ans/2);
}

int main() {
    scanf("%d",&n); // 输入点的个数
    for (int i=1;i<=n;++i) {
        Point P,Q;double a,b,c,d; // 输入四个点的坐标 x1,y1,x2,y2
        scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
        P=Point(a,b);Q=Point(c,d); // 得到点
        L[i]=Line(P,Q-P); // 通过一个点和线得到 直线
    } // 开始需要设置一个无限大的区域
    L[++n]=Line(Point(0,10000),Vector(0,-10000));
    L[++n]=Line(Point(0,0),Vector(10000,0));
    L[++n]=Line(Point(10000,0),Vector(0,10000));
    L[++n]=Line(Point(10000,10000),Vector(-10000,0));
    halfp();
    printf("%.1lf\n",Area());
    return 0;
}