C++(clang++11) 解法, 执行用时: 7ms, 内存消耗: 512K, 提交时间: 2021-04-25 03:28:24

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-6;

struct point{
	double x,y; 
	point(){}
	point(double X,double Y){
		x=X;y=Y;
	}
};

struct line{
	point a,b;
	line(point X,point Y){
		a=X;b=Y;
	}
};

double xmult(point p1,point p2,point p3){return (p1.x-p3.x)*(p2.y-p3.y)-(p2.x-p3.x)*(p1.y-p3.y);}

double area(point p1,point p2,point p3){return fabs(xmult(p1,p2,p3))/2;}

double dot_mul(point p1,point p2,point p0){return (p1.x-p0.x) * (p2.y-p0.y) - (p2.x-p0.x) * (p1.y-p0.y);}

bool dot_online_in(point a,line b){return fabs(dot_mul(a,b.a,b.b))<eps&&a.x>=min(b.a.x,b.b.x)&&a.x<=max(b.a.x,b.b.x)&&a.y>=min(b.a.y,b.b.y)&&a.y<=max(b.a.y,b.b.y)!=0;}

point intersection(line u,line v){
	point ret=u.a;
	double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
	ret.x+=(u.b.x-u.a.x)*t;
	ret.y+=(u.b.y-u.a.y)*t;
	return ret;
}


int main(){
	point a,b,c,d,e;
	double x,y;
	cin>>x>>y; a={x,y};
	cin>>x>>y; b={x,y};
	cin>>x>>y; c={x,y};
	
	double areaSum=area(a,b,c);
	
	cin>>x>>y; d=point(x,y);
	cin>>x>>y; e=point(x,y);
	line l={d,e};
	
	if(dot_online_in(a,l)){
		point f=intersection(l,line(b,c));
		double area2=area(f,a,c);
		printf("%.10lf\n",min(areaSum-area2,area2));
	}
	else if(dot_online_in(b,l)){
		point f=intersection(l,line(a,c));
		double area2=area(f,b,c);
		printf("%.10lf\n",min(areaSum-area2,area2));
	}
	else if(dot_online_in(c,l)){
		point f=intersection(l,line(a,b));
		double area2=area(f,c,a);
		printf("%.10lf\n",min(areaSum-area2,area2));
	}
	else{
		double area2;
		if(xmult(a,l.a,l.b)*xmult(b,l.a,l.b)>eps){
            point xx1=intersection(l,line(a,c));
            point xx2=intersection(l,line(b,c));
            area2=area(xx1,xx2,c);
		}
		else if(xmult(b,l.a,l.b)*xmult(c,l.a,l.b)>eps){
            point xx1=intersection(l,line(a,c));
            point xx2=intersection(l,line(a,b));
            area2=area(xx1,xx2,a);
		}
		else if(xmult(c,l.a,l.b)*xmult(a,l.a,l.b)>eps){
            point xx1=intersection(l,line(a,b));
            point xx2=intersection(l,line(b,c));
            area2=area(xx1,xx2,b);
		}
		printf("%.10lf\n",min(areaSum-area2,area2));
	}
	return 0;
}