POJ 3130 How I Mathematician Wonder What You Are!

POJ_3130

    对于半平面交的一些简明扼要的介绍可以参考这篇博客：http://blog.csdn.net/accry/article/details/6070621。此外，这篇博客上介绍的还有我敲出的程序都只是比较好理解的O(n^2)的求半平面交的算法，对于O(nlogn)的算法可以参考朱泽园的论文。

    由于这个题目指明了多边形上的点是按逆时针序给出的，因而就不用再将每组数据都其统一成某个顺序了。

#include<stdio.h>
#include<string.h>
#define MAXD 110
#define zero 1e-8
#define INF 100000
struct point
{
    double x, y;
}p[MAXD], wa[MAXD], wb[MAXD], *a, *b;
int N, an, bn;
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;
    for(i = 0; i < N; i ++)
        scanf("%lf%lf", &p[i].x, &p[i].y);
    p[N] = p[0];
}
void add(double x, double y)
{
    b[bn].x = x, b[bn].y = y;
    ++ bn;
}
void cut(int k)
{
    int i, j, tn;
    point *t;
    double x, y, t1, t2;
    bn = 0;
    for(i = 0; i < an; i ++)
    {
        t1 = det(p[k + 1].x - p[k].x, p[k + 1].y - p[k].y, a[i].x - p[k].x, a[i].y - p[k].y);
        t2 = det(p[k + 1].x - p[k].x, p[k + 1].y - p[k].y, a[i + 1].x - p[k].x, a[i + 1].y - p[k].y);
        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;
    tn = an, an = bn, bn = tn;
    a[an] = a[0];
}
void solve()
{
    int i, j, k;
    a = wa, b = wb;
    an = 4;
    a[0].x = -INF, a[0].y = -INF, a[1].x = INF, a[1].y = -INF, a[2].x = INF, a[2].y = INF, a[3].x = -INF, a[3].y = INF;
    a[4] = a[0];
    for(i = 0; i < N; i ++)
        cut(i);
    if(an == 0)
        printf("0\n");
    else
        printf("1\n");
}
int main()
{
    for(;;)
    {
        scanf("%d", &N);
        if(!N)
            break;
        init();
        solve();
    }
    return 0;
}