The Way to a Friends House Is Never Too Long
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You are given a set of points in the 2D plane. You start at the point with the least X and greatest Y value, and end at the point with the greatest X and least Y value. The rule for movement is that you can not move to a point with a lesser X value as compared to the X value of the point you are on. Also for points having the same X value, you need to visit the point with the greatest Y value before visiting the next point with the same X value. So, if there are 2 points: (0,4 and 4,0) we would start with (0,4) - i.e. least X takes precedence over greatest Y. You need to visit every point in the plane.
You will be given an integer t(1<=t<=20) representing the number of test cases. A new line follows; after which the t test cases are given. Each test case starts with a blank line followed by an integer n(2<=n<=100000), which represents the number of points to follow. This is followed by a new line. Then follow the n points, each being a pair of integers separated by a single space; followed by a new line. The X and Y coordinates of each point will be between 0 and 10000 both inclusive.
For each test case, print the total distance traveled by you from start to finish; keeping in mind the rules mentioned above, correct to 2 decimal places. The result for each test case must be on a new line.
Input: 3 2 0 0 0 1 3 0 0 1 1 2 2 4 0 0 1 10 1 5 2 2 Output: 1.00 2.83 18.21 For the third test case above, the following is the path you must take: 0,0 -> 1,10 1,10 -> 1,5 1,5 -> 2,2 = 18.21
|Time Limit:||1.67667 - 3.67667 sec|
|Source Limit:||50000 Bytes|
|Languages:||C, CPP14, JAVA, PYTH, PYTH 3.6, PYPY, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, rust, SCALA, swift, D, PERL, FORT, ADA, CAML, ASM, BASH, LISP sbcl, LISP clisp, JS, ERL, TCL, kotlin, PERL6, TEXT, SCM chicken, PYP3, CLOJ, R, COB, FS|
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