JOJ 2676 Problem B

#include<stdio.h>
#include<math.h>
struct POINT
{
    double x,y;
}node[4];

double det(POINT p1,POINT p2, POINT p3) //p1p2*p1p3
{
    return (p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y);
}

int main()
{
    while(scanf("%lf",&node[1].x)!=EOF)
    {
        scanf("%lf",&node[1].y);
        for (int i=2;i<=3;i++)
           scanf("%lf%lf",&node[i].x,&node[i].y);
        double s=fabs(1/2.0*det(node[1],node[2],node[3]));
        double m1,m2;
        if(det(node[1],node[2],node[3])>0)//2在3的逆时针
        {
            m1=s+node[3].x*node[2].y-node[2].x*node[3].y;
            m2=1/3.0*s+node[2].x*node[1].y-node[1].x*node[2].y;
        }
        else//顺时针，枚举所有情况不难发现这个正负规律
        {
            m1=-s+node[3].x*node[2].y-node[2].x*node[3].y;
            m2=-1/3.0*s+node[2].x*node[1].y-node[1].x*node[2].y;
        }
        double t1=node[3].x-node[2].x;
        double t2=node[2].y-node[3].y;
        double t3=node[2].x-node[1].x;
        double t4=node[1].y-node[2].y;
        double x=(m1*t3-m2*t1)/(t2*t3-t1*t4);
        double y=(t4*m1-t2*m2)/(t1*t4-t2*t3);
        printf("%.3lf %.3lf\n",x,y);
    }
    return 0;
}