Lightoj-1433-minimum ARC Distance (geometry)

1433-minimum ARC Distance
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Time Limit:2 second (s) Memory limit:32 MB
All probably know how to calculate the distance between both points in and dimensional Cartesian plane. Problem you have to find the minimum arc distance between the points and they is on a circle centered at Anot Her point.

You'll be given the co-ordinates of the points A and B and co-ordinate of the center O. You just has to calculate the minimum arc distance between A and B. In the picture, you had to calculate the length of the arc ACB. You can assume that A and B would always be on the circle centered at O.

Input
Input starts with an integer T (≤100), denoting the number of test cases.

Each case starts with a line containing six integers ox, Oy, Ax, Ay, Bx, by where (Ox, Oy) indicates the co-ordinate of O, (Ax, Ay) denote the co-ordinate of A and (Bx, by) denote the co-ordinate of B. All the integers would lie in the range [1, 10000].

Output
For each case, print the case number and the minimum arc distance. Errors less than 10-3 'll be ignored.

Sample Input
Output for Sample Input
5
5711 3044 477 2186 3257 7746
3233 31 3336 1489 1775 134
453 4480 1137 6678 2395 5716
8757 2995 4807 8660 2294 5429
4439 4272 1366 8741 6820 9145
Case 1:6641.81699183
Case 2:2295.9288
Case 3:1616.690325
Case 4:4155.6415934
Case 5:5732.01250253

#include <iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespacestd;Const DoublePI =3.1415926535898;DoubleDistDoubleX1,DoubleY1,DoubleX2,Doubley2) {    returnsqrt (Pow (x1-x2,2) +pow (Y1-y2,2));}intMain () {intT; DoubleOx,oy,ax,ay,bx,by,a,b,c,arc; scanf ("%d",&T);  for(intt=1; t<=t;t++) {scanf ("%LF%LF%LF%LF%LF%LF",&ox,&oy,&ax,&ay,&bx,&y); A=Dist (Ox,oy,ax,ay); C=Dist (ax,ay,bx,by); Arc= ACOs ((a*a*2-C*C)/(2*a*a)); printf ("Case %d:%.3lf\n", t,a*arc); }        return 0;} 

Lightoj-1433-minimum ARC Distance (geometry)