Computational Geometry--bnu51638

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Bored river boat classmates and bored fish swell students are very fond of playing table hockey (actually just like listening to the sound of the ball collision). In the boring day, the bored river boat students think of a boring play: two people at the same time two balls on the table, while hitting out, and then listen to the two balls hit together when the sound. However, they are not very good at the accuracy of the shot, so the two balls will not be able to collide.

Now assume that the desktop is infinitely large and absolutely smooth, giving the initial position, radius, and movement speed of the two balls, ensuring that the two balls are initially untouched. Bored river boat students want to know whether two goals can collide (contact is think of collision), if can, he wants to know two ball collision time (from two people to start timing), if not, he would like to know the whole process of the minimum distance between two balls, here the two ball distance is defined as two balls take two points distance of the minimum value, The data guarantees that in this case the answer is not less than. Note that the puck is a cylinder, which looks down from the air as a circle, and in this problem, the height of the puck is negligible. Input

The first line is a positive integer that represents the number of groups of test data,

Each set of test data contains two rows,

The first line contains five integers that are not greater than the absolute value, representing the initial position, radius, and motion velocity of the first ball. Output

For each set of test data, if two balls can collide, output two ball collision time, otherwise the output of the whole process of the two ball distance of the minimum, relative error does not exceed, that is, the output is, the standard answer is, if satisfied, the output will be considered the correct answer. Sample Input

2
0 0 2 1 0 each
0 1-1 0
0 0 2 1 0
11 5 1-1 0

Sample Output

4.0000000000
2.0000000000

Hint

For the first set of samples, the two balls collide 4 seconds after the hit,

For the second set of samples, the two balls do not collide and the two balls are closest to each other 5.5 seconds after hitting the ball, and the distance is 2.0. Source 14th Beijing Normal University Program design competition final Author quailty Solution:

Set time is T two ball collision or two ball distance nearest, with T can represent two ball coordinates Q1 (x1+t*vx1, Y1+t*vy1) Q2 (x2+t*vx2, Y2+t*vy2),

Q1Q2 distance = aqrt{(x1-x2) ^2 + (y1-y2) ^2} =r1+r2; The collation equation gets a unary two-time equation about T sqrt (a t^2 + b T + c) = R1+R2;

Solve the equation, divide a=0 and a. =0,a.   =0, solving <0, =0, >0, when <0 without solution, solve the equation to the left of the minimum value, that is, T =-b/(2*a), t>0, bring into the equation d= sqrt ((4ac-b^2)/(4a)), ans=d-(R1+R2); T<0, Ans=sqrt (c)-(R1+R2). When >=0, Ans=min (T1,T2). When A=0, Ans=sqrt (c)-(R1+R2).

Also reviewed the middle school knowledge

#include <iostream> #include <stdio.h> #include <math.h> #include <string.h> using namespace std
;
    Double Solve (double a,double b,double c) {double dat=b*b-4.0*a*c;
    if (dat<0) return-1;
    Double x1= (-b+sqrt (DAT))/2.0/a;
     Double x2= (-b-sqrt (DAT))/2.0/a;
    if (x2>=0.0) return x2;
    if (x2<0&&x1>0) return x1;
if (x1<0&&x2<0) return-1;
    } int main () {int t;
    Double x1,y1,r1,vx1,vy1;
    Double x2,y2,r2,vx2,vy2;
    Double X,vx,y,vy;
    Double a,b,c;
    Double R;
    scanf ("%d", &t);
        while (t--) {scanf ("%lf%lf%lf%lf%lf", &x1,&y1,&r1,&vx1,&vy1);
        scanf ("%lf%lf%lf%lf%lf", &x2,&y2,&r2,&vx2,&vy2);
        x=x1-x2;///c y=y1-y2;///c vx=vx1-vx2;///a vy=vy1-vy2;///a a=vx*vx+vy*vy;
        C=x*x+y*y;
        b=2.0* (X*vx+y*vy);
        r=1.0* (R1+R2) * (R1+R2); Double t=-1.0*b/2.0/A;
            if (a!=0) {double ans=solve (a,b,c-r);

            if (ans<0&&t<0) printf ("%.10lf\n", sqrt (c)-r1-r2);

            if (ans<0&&t>=0) printf ("%.10lf\n", sqrt ((4.0*a*c-b*b)/(4.0*a))-R1-R2);
        if (ans>=0) printf ("%.10lf\n", ans);
    } else printf ("%.10lf\n", sqrt (c)-r1-r2);
} return 0;
 }