90 lines
2.8 KiB
C++
90 lines
2.8 KiB
C++
#include<cstdio>
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#include<iostream>
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#include<complex>
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#include<cmath>
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using namespace std;
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#define Y imag
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#define X real
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const double eps = 1e-9;
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const char *err = "Error";
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const double pi = acos(-1.0);
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typedef complex<double> pnt;
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static double dot(const pnt &a, const pnt& b) {return X(conj(a)*b);}
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static double cross(const pnt &a, const pnt &b) {return imag(conj(a)*b);}
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void work(const pnt& s, double argu) {
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while(argu<0) argu+=2*pi;
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while(argu >= 2*pi) argu -=2*pi;
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if(argu < pi+eps) {
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puts(err);
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return ;
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}
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argu = - argu;
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printf("%.3lf\n", X(s) +( abs(argu + 3*pi/2)<eps ? 0 : Y(s)/tan(argu) ));
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}
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pnt jdg(const pnt& p0, const pnt& p1,const pnt& p3, const pnt& p2){
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if(cross(p3-p0,p1-p0) * cross(p2-p0,p1-p0) > -eps) return p0;
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double s1 = cross(p2-p1,p0-p1), s2 = cross(p3-p1,p0-p1);
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s1 = abs(s1), s2 = abs(s2);
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double x = (X(p2) *s2 + X(p3)*s1) / (s1+ s2);
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double y = (Y(p2) *s2 + Y(p3)*s1) / (s1+ s2);
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pnt p(x,y);
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if(cross(p3-p2,p1-p2)*cross(p3-p2,p0-p2)>0 && abs(p-p0) < abs(p-p1)) return p0;
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return p;
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}
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double cal(double arg0, double arg1, double tmp) {
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double t1 =arg1+pi/2-(arg0+pi);
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double s1 = sin(t1);
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double s2 = s1/tmp;
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if(s2>=1.0 || s2 <=-1.0) return 1e10;
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else return arg1-pi/2-asin(s2);
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}
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int main(){
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int tst;
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double tmp;
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cin >> tst;
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pnt p[3];
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while(tst--) {
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double x1,y1,x2,y2;
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while(cin >> x1 >> y1 >> x2 >> y2){
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pnt p1(x1,y1), p2(x2,y2), p0;
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int s = -1;
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double mx = 1e10;
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for(int i=0;i<3;i++){
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cin >> x1 >> y1;
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p[i] = pnt(x1,y1);
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}
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cin >> tmp;
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for(int i=0;i<3;i++) {
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pnt P = jdg(p1,p2,p[i],p[(i+1)%3]);
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if(P == p1) continue;
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double v = abs(p1-P);
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if(v < mx) mx = v, s = i, p0 = P;
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}
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if(s == -1){
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work(p1, arg(p2-p1));
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continue;
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}
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pnt p3;
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double t = 0;
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if(cross(p2-p1, p[(s+1)%3] - p[s])<0)t = pi;
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double arg0 = cal(arg(p2-p1),arg(p[(s+1)%3] - p[s])+t,tmp);
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pnt np = pnt(X(p0) + 200*cos(arg0), Y(p0) + 200*sin(arg0));
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int k = -1;
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for(int i=0;i<3;i++) {
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if(i==s) continue;
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pnt P = jdg(p0,np,p[i],p[(i+1)%3]);
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if(P == p0) continue;
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k = i;p3 = P;
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}
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t = 0;
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if(cross(p3-p0, p[(k+1)%3] - p[k])<0)t = pi;
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double arg1 = cal(arg(p3 - p0), arg(p[(k+1)%3] - p[k])+t, 1.0/tmp);
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if(arg1 == 1e10) {
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puts(err);
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continue;
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}
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work(p3,arg1);
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}
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}
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}
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