CSU 1407: Shortest distance (math AH)

Title Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1407

Descriptiontwo points A,  Bare doing uniform motion. Give the T= 0 Moments A,  Bthe coordinates, and A,  Bthe speed, calculation T≥0 O'Clock the minimum value of the distance between two points. Inputthe first line of input contains an integer T(1≤ T≤ $), indicating that there are altogether TGroup test Data. for each set of test data, the first row consists of 4 integers x A,  y A,  v Ax,  v Ay( -103≤ x A,  y A,  v Ax,  v Ay≤Ten3), indicating T= 0 Moments Athe coordinates are ( x A,  y A), Athe speed in xthe component in the axis direction is v Ax, in ythe component on the axis is v Ay. The second line consists of four integers x B,  y B,  v Bx,  v by( -103≤ x B,  y B,  v Bx,  v by≤Ten3), in the same way given the Bof the various properties. Outputfor each set of test data, the output T ≥0 O'Clock the minimum value of the distance between two points, preserving 8 decimal places. Sample Input

60 0 0 00 1 0 10 0-1 10 0 1-10 1 1 02 0 0 10 1 1 02 0 1 00 0-1 11 1 1-1997 997-1000-1000-1000-1000 1000 1000

Sample Output

1.000000000.000000000.707106782.236067981.414213560.00000000

HINT

Source

The eighth annual college student Program Design Competition

The code is as follows:

#include <cstdio> #include <cmath> #include <cstring>int main () {int t;    Double xa, ya, vax, Vay;    Double XB, YB, VBX, Vby;    scanf ("%d", &t);        while (t--) {scanf ("%lf%lf%lf%lf", &xa,&ya,&vax,&vay);        scanf ("%lf%lf%lf%lf", &xb,&yb,&vbx,&vby); dis^2 = A*t^2+b*t+c Double A = ((VAX-VBX) * (VAX-VBX) + (vay-vby) * (Vay-vby));        A must be >=0 double b = xa* (VAX-VBX) +xb* (Vbx-vax) +ya* (Vay-vby) +yb* (Vby-vay));        Double c = xa* (XA-2*XB) +ya* (YA-2*YB) +yb*yb+xb*xb;        Double dis1 = 0, Dis2 = 0, dis3 = 0;        Double xx =-(2*a)/b;//vertex coordinates double yy = sqrt ((4*a*c-b*b)/(4*a));        Double d =-b/(2*a);//Solution: derivative: 2*a*x+b = 0;            if (a = = 0) {dis1 = sqrt (c);        printf ("%.8lf\n", dis1);            } else if (a > 0) {if (d >= 0) {dis2 = yy;          } else {dis2 = sqrt (c);  } printf ("%.8lf\n", Dis2);    }//printf ("dis1:%.8lf\n", dis1);//printf ("dis2:%.8lf\n", Dis2); } return 0;}

CSU 1407: Shortest distance (math AH)