POJ 2540 Hotter Colder

POJ_2540

    不妨设上一个点为p，现在走到的点为c，显然到p的距离和到c的距离相等的点就在线段pc的中垂线上，而这条直线把平面分成了两个半平面，如果是Hotter就说明object在更靠近c的这个半平面内，如果是Colder就说明object在更靠近p的这个半平面，于是，我们就把这个问题转化成半平面交的问题了。

#include<stdio.h>
#include<string.h>
#define MAXD 210
#define INF 10
#define zero 1e-8
struct point
{
    double x, y;
}wa[MAXD], wb[MAXD], *a, *b;
int N, na, nb;
double px, py, cx, cy;
char st[10];
double fabs(double x)
{
    return x < 0 ? -x : x;
}
int dcmp(double x)
{
    return fabs(x) < zero ? 0 : (x < 0 ? -1 : 1);
}
double det(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}
void init()
{
    int i, j, k;
    a = wa, b = wb;
    a[0].x = 0, a[0].y = 0, a[1].x = INF, a[1].y = 0, a[2].x = INF, a[2].y = INF, a[3].x = 0, a[3].y = INF;
    a[na = 4] = a[0];
    px = py = 0;
}
void add(double x, double y)
{
    b[nb].x = x, b[nb].y = y;
    ++ nb;
}
void cut(double x0, double y0, double nx, double ny)
{
    int i, j, k;
    point *t;
    double t1, t2, x, y;
    nb = 0;
    for(i = 0; i < na; i ++)
    {
        t1 = det(nx, ny, a[i].x - x0, a[i].y - y0);
        t2 = det(nx, ny, a[i + 1].x - x0, a[i + 1].y - y0);
        if(dcmp(t1) >= 0)
            add(a[i].x, a[i].y);
        if(dcmp(t1) * dcmp(t2) < 0)
        {
            x = (fabs(t2) * a[i].x + fabs(t1) * a[i + 1].x) / (fabs(t1) + fabs(t2));
            y = (fabs(t2) * a[i].y + fabs(t1) * a[i + 1].y) / (fabs(t1) + fabs(t2));
            add(x, y);
        }
    }
    t = a, a = b, b = t;
    a[na = nb] = a[0];
}
void calculate()
{
    int i, j, k;
    double s = 0;
    for(i = 0; i < na; i ++)
        s += det(a[i].x, a[i].y, a[i + 1].x, a[i + 1].y);
    printf("%.2lf\n", fabs(s) / 2);
}
void solve()
{
    int i, j, k;
    if(st[0] == 'S')
        na = 0;
    else if(st[0] == 'H')
        cut((px + cx) / 2, (py + cy) / 2, cy - py, px - cx);
    else
        cut((px + cx) / 2, (py + cy) / 2, py - cy, cx - px);
    calculate();
}
int main()
{
    init();
    while(scanf("%lf%lf%s", &cx, &cy, st) == 3)
    {
        solve();
        px = cx, py = cy;
    }
    return 0;
}