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Home » Practice(easy) » Obstacle Course

Obstacle Course

Problem code: COURSE

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All submissions for this problem are available.

A number of traffic cones have been placed on a circular racetrack to form an obstacle course. You are asked to determine the largest sized car that can navigate the course. For simplicity, the cones are assumed to have zero width and the car is perfectly circular and infinitely maneuverable. The track itself is the area between 2 concentric circles.

Formally, the course can be navigated by a car of radius c if there exists a closed loop around the center of the track which lies between the circles forming the track, and every point on the loop is at least c distance away from each cone and each boundary of the track.

Input:

Input begins with an integer T (about 25), the number of test cases. Each test case begins with 2 integers r and R (1<=r<R<=25000). The racetrack is the area between the circles centered at (0,0) with radii r and R. The next line of input contains an integer N (0<=N<=500), the number of cones. N lines follow, each containing the coordinates of a cone. The coordinates are integers, and are guaranteed to lie within the track, and be distinct. Cases are separated by a blank line.

Output:

For each input, output on a single line the diameter of the largest car that can navigate the course, rounded to 3 decimal places.

Sample input:

1

5 10
3
6 0
5 7
-2 -7

Sample output:

2.720

Explanation:

The image below shows the course corresponding to the sample input. The black circles represent the boundaries of the racetrack, the small dots the cones, and the filled red circle the car. Also shown is one possible path of the car through the course.

sample course

Author: pieguy
Date Added: 9-06-2010
Time Limit: 1 sec
Source Limit: 50000 Bytes
Languages: ADA, ASM, BASH, BF, C, C99 strict, CAML, CLOJ, CLPS, CPP 4.0.0-8, CPP 4.3.2, CS2, D, ERL, F#, FORT, GO, HASK, ICK, ICON, JAR, JAVA, JS, LISP clisp, LISP sbcl, LUA, NEM, NICE, PAS fpc, PAS gpc, PERL, PERL6, PHP, PIKE, PRLG, PYTH, PYTH 3.1.2, RUBY, SCALA, SCM guile, SCM qobi, ST, TEXT, WSPC


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Comments

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bounce!!!

ritwick @ 13 Aug 2010 11:00 PM

bounce!!!

i m getting rite ans on my

akash_d_learner @ 21 Aug 2010 04:04 AM

i m getting rite ans on my comp for this prob but while submitting i m getting wrong ans. pls help me to figure whats wrongin my code..

 

#include <cstdio>
#include <math.h>
using namespace std;
int main()
{
int t,r,rr,n,*a,*b;
float d,l1,l2,ln,le=25000;
scanf("%d",&t);
for(int i=0;i<t;i++)
{
scanf("%d%d%d",&r,&rr,&n);
a=new int [n];
b=new int [n];
for(int j=0;j<n;j++)
{
scanf("%d%d",&a[j],&b[j]);
d=sqrt(a[j]*a[j]+b[j]*b[j]);
l1=d-r;
l2=rr-d;
if (l1>l2)
ln=l1;
else
ln=l2;
if(le>ln)
le=ln;
for(int k=0;k<j;k++)
{
l2=sqrt(((a[j]-a[k])*(a[j]-a[k]))+((b[j]-b[k])*(b[j]-b[k])));
if(le>l2)
le=l2;
}
}
printf("%5.3fn",le);
}
}

 

thanks in advance.

can anyone explain what we

mukulbudania @ 1 Sep 2010 10:33 PM

can anyone explain what we are supposed to find?? i am not even able to get the question?? Thanks in advance..

u need to find the diameter

jyotigupta2203 @ 18 Oct 2010 12:09 AM

u need to find the diameter of the largest car that can pass through the area between the concentric circles.

the point of the obstacles are given .so the diameter can vary from r circle boundary to point (on line x=y) or from point to R circle boundary ..

can anyone please explain me

tarun_kumar @ 23 Nov 2010 07:29 PM

can anyone please explain me the application of prim's algorithm in solving this problem.

i know prim's algorithm but not getting its application in one of the solutions.

Can we modify the way input

meher_almighty @ 2 Aug 2011 09:56 PM
Can we modify the way input is taken? or is it have to be in the exact same way given above?

hello friends.... the algo

abhipranay @ 14 Dec 2011 02:24 PM
hello friends.... the algo which i am following seems right to me to solve this problem but i am getting wrong answer on submission. for sample input my answer is correct.

help plz!..I am using the

jigsaw004 @ 19 May 2012 01:52 AM
help plz!..I am using the following algorithm: calculate the distance of (x,y) from origin(say d); if it is greater than (R+r/2), then check whether (d-r) is less than max, (which was initially set to R-r). If yes(i.e. if (d-r) is less than max), then update max to (d-r). If d<((R+r)/2) then check whether (R-d) is less than max. If yes, then update max to (R-d). Is anything wrong in this algorithm? IF yes then please let me know!

at any obstacle, their is one

jigsaw004 @ 19 May 2012 01:55 AM
at any obstacle, their is one wider path and other narrow. We always have the authority to chose the wider path. But, the cars diameter should be equal to the narrowest of these 'wider' paths...This logic seems correct. However, always getting wrong answer!..

SUCCESSFUL SUBMISSIONS FOR THIS PROBLEM:

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Program should read from standard input and write to standard output. After you submit a solution you can see your results by clicking on the [My Submissions] tab on the problem page. Below are the possible results:

 

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