POJ 1329 Circle Through Three Points

POJ_1329

    题目主要就是在求三角形的外心，即各边中垂线的交点，推导出公式后直接计算即可。

#include<stdio.h>
#include<string.h>
#include<math.h>
#define zero 1e-8
double x3, y3, x4, y4, x5, y5;
double dcmp(double x)
{
    if(fabs(x) < zero)
        return 0;
    if(x < 0)
        return -1;
    return 1;
}
char sign(double x)
{
    return x < 0 ? '-' : '+';
}
void solve()
{
    double x1, y1, x2, y2, k1, k2, x, y, r, a1, a2, a3;
    if(dcmp(y3 - y4) == 0)
    {
        x1 = (x3 + x4) / 2, y1 = y3;
        k2 = (x4 - x5) / (y5 - y4);
        x2 = (x4 + x5) / 2, y2 = (y4 + y5) / 2;
        x = x1, y = k2 * (x1 - x2) + y2;
    }
    else
    {
        if(dcmp(y4 - y5) == 0)
        {
            x2 = (x4 + x5) / 2, y2 = y4;
            k1 = (x3 - x4) / (y4 - y3);
            x1 = (x3 + x4) / 2, y1 = (y3 + y4) / 2;
            x = x2, y = k1 * (x2 - x1) + y1;
        }
        else
        {
            x1 = (x3 + x4) / 2, y1 = (y3 + y4) / 2;
            k1 = (x3 - x4) / (y4 - y3);
            x2 = (x4 + x5) / 2, y2 = (y4 + y5) / 2;
            k2 = (x4 - x5) / (y5 - y4);
            x = (y2 - y1 + k1 * x1 - k2 * x2) / (k1 - k2);
            y = k1 * (x - x1) + y1;
        }
    }
    r = sqrt((x - x3) * (x - x3) + (y - y3) * (y - y3));
    printf("(x %c %.3lf)^2 + (y %c %.3lf)^2 = %.3lf^2\n", sign(-x), fabs(x), sign(-y), fabs(y), r);
    a1 = (-2) * x, a2 = (-2) * y, a3 = x * x + y * y - r * r;
    printf("x^2 + y^2 %c %.3lfx %c %.3lfy %c %.3lf = 0\n", sign(a1), fabs(a1), sign(a2), fabs(a2), sign(a3), fabs(a3));
}
int main()
{
    while(scanf("%lf%lf%lf%lf%lf%lf", &x3, &y3, &x4, &y4, &x5, &y5) == 6)
    {
        solve();
        printf("\n");
    }
    return 0;
}