This Takes the Cake 作呕的面积剖分 不过是水题

This Takes the Cake 恶心的面积剖分 不过是水题

#include <iostream>
#include <cmath>
using namespace std;
struct point
{
    double x, y;
};
double a1, a2, d, s;
double area(int n, point* p)
{
    double s1=0, s2=0;
    for(int i=0; i<n; i++) s1 += p[(i+1)%n].y*p[i].x, s2 += p[(i+1)%n].y*p[(i+2)%n].x;
    return fabs(s1-s2)/2;
}
void check(double s1)
{
    double s2 = s - s1;
    double ds = fabs(s1 - s2);
    if(ds < d)
    {
        if(s1 > s2) swap(s1, s2);
        d = ds, a1 = s1, a2 = s2;
    }
}
void solve(point *p)
{
    point p1[4], p2[4];
    int i;
    for(i=0; i<4; i++) p1[i].x = (p[i].x+p[(i+1)%4].x)/2, p1[i].y = (p[i].y+p[(i+1)%4].y)/2;
    for(i=0; i<4; i++)
    {
        p2[0] = p[i], p2[1] = p[(i+1)%4], p2[2] = p[(i+2)%4];
        check(area(3, p2));
        p2[2].x = (p[(i+1)%4].x+p[(i+2)%4].x)/2, p2[2].y = (p[(i+1)%4].y+p[(i+2)%4].y)/2;
        check(area(3, p2));
        p2[2] = p[(i+3)%4], p2[1].x = (p[(i+2)%4].x+p[(i+3)%4].x)/2, p2[1].y = (p[(i+2)%4].y+p[(i+3)%4].y)/2;
        check(area(3, p2));
        p2[1] = p[i+1], p2[2] = p1[i+1], p2[3] = p1[(i+3)%4];
        check(area(4, p2));
        p2[0] = p1[i], p2[1] = p[(i+1)%4], p2[2] = p1[(i+1)%4];
        check(area(3, p2));
    }
}
int main()
{
    point p[4];
    int i, j=0;
    while(1)
    {
        d = 1000000;
        for(i=0; i<4; i++) cin >> p[i].x >> p[i].y;
        if(!p[0].x && !p[0].y && !p[1].x && !p[1].y) break;
        s = area(4, p);
        solve(p);
        printf("Cake %d: %.3lf %.3lf\n", ++j, a1, a2);
    }
}