Dwarven Sniper’s hunting

Now the hunting starts in the world named DOTA, a stupid PC game which cannot play with others together.
Among the individuals in the game, there are two heroes named Dwarven Sniper and Lycanthrope. Lycanthrope wants to escape from being captured; however, our Dwarven Sniper won’t let him go! He will use the Silver Bullet to kill the Lycanthrope by only one shot! Yes, that’s enough.
  Lycanthrope is running on a line in the map with a constant speed and direction. The weapon range of the Silver Bullet is limited by L meters. Dwarven Sniper can run for a while freely, and then shoot Lycanthrope. In order to show his excellent shooting skill, Dwarven Sniper wants the Silver Bullet flying as far as possible. But don’t forget the flying time of the Silver Bullet due to considerable weight of the bullet. And Dwarven Sniper wants to stop the hunting as quickly as possible. So if there is more than one way to show his excellent skill, he would choose the fastest way. In this problem we consider the Silver Bullet and Lycanthrope as two points.
Now Dwarven Sniper wants to know the maximum length that the Silver Bullet can fly, and the shortest time that the hunting lasts. Specifically, the total hunting time is defined as the time interval from the start of hunting to the moment that the bullet hit Lycanthrope. Can you help him?

-1 0 0 10 1 0 2 10 10
0 0 0 5 0 1 2 6 6
0 0 0 0 0 0 0 0 0

10.000 1.000
6.000 3.000

题目大意：

D追杀L，D是一个远程英雄，D可以发出距离为L的技能，已知L这个英雄的起始位置为X1,Y1,移动速度方向矢量是LX,LY，D的起始位置为X2,Y2，D的移动速度是VD，D发出L距离的弓箭的移动速度是VB，(Lx*Lx+Ly*Ly) < vD*vD < vB*vB，问你在D充分表现自己射击能力的情况下，最少多长时间杀死L?

因为：(Lx*Lx+Ly*Ly) < vD*vD < vB*vB，也就是说：L的移动速度<D的移动速度<弓箭的移动速度。也就是D表现自己的能力的话，一定射击的最长距离为L，那么时间怎么算，可以二分出答案。

解题代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

// X1, Y1, X2, Y2, Lx, Ly, vD , vB and L

const double eps=1e-7;
double X1,Y1,X2,Y2,lx,ly,vd,vb,L;

double getdis(double ax,double ay,double bx,double by){
    return sqrt ( (ax-bx)*(ax-bx)+(ay-by)*(ay-by) );
}

double getmin(double ans1,double ans2){
    if(ans1<-eps) return ans2;
    if(ans2<-eps) return ans1;
    return min(ans1,ans2);
}

void solve(){
    double l=L/vb,r=1e9;
    while(r-l>eps){
        double mid=(l+r)/2;
        double X0=X1+mid*lx,Y0=Y1+mid*ly;
        double dis0=getdis(X0,Y0,X2,Y2);
        double disd=vd*(mid-L/vb);
        if(dis0<=L){
            if(disd+dis0<=L) l=mid;
            else r=mid;
        }else{
            if(disd+L>=dis0) r=mid;
            else l=mid;
        }
    }
    printf("%.3lf %.3lf\n",L,(r+l)/2);
}

int main(){
    while(cin>>X1>>Y1>>X2>>Y2>>lx>>ly>>vd>>vb>>L){
        if( fabs(X1)<eps && fabs(X2)<eps && fabs(Y1)<eps && fabs(Y2)<eps ){
            if( fabs(lx)<eps && fabs(ly)<eps && fabs(vd)<eps && fabs(vb)<eps && fabs(L)<eps ) break;
        }
        solve();
    }
    return 0;
}