double rotating_calipers(const point *ps, int n)
{
    double res = 0;
    int i, j, k;
    j = 1; k = 2;
    for(i=0; i<n; i++)
    {
        while((xmult(ps[j], ps[k], ps[i]) < xmult(ps[j], ps[(k+1)%n], ps[i])) || (xmult(ps[j], ps[k], ps[i]) < xmult(ps[(j+1)%n], ps[k], ps[i]))
             )
        {
        while(xmult(ps[j], ps[k], ps[i]) < xmult(ps[j], ps[(k+1)%n], ps[i]))
            k = (k+1) % n;
        while(xmult(ps[j], ps[k], ps[i]) < xmult(ps[(j+1)%n], ps[k], ps[i]))
            j = (j+1) % n;
        }
        res = max(res, xmult(ps[j], ps[k], ps[i])/2.0);
    }
    return res;
}


这么写能过你的数据..但是感觉复杂度其实不高?? 想不到证明的办法..求解答