MCOCIR - Cocircular Points
You probably know what a set of collinear points is: a set of points such that there exists a straight line that passes through all of them. A set of cocircular points is defined in the same fashion, but instead of a straight line, we ask that there is a circle such that every point of the set lies over its perimeter.
The International Collinear Points Centre (ICPC) has assigned you the following task: given a set of points, calculate the size of the larger subset of cocircular points.
Input
Each test case is given using several lines. The first line contains an integer N representing the number of points in the set (1 ≤ N ≤ 100). Each of the next N lines contains two integers X and Y representing the coordinates of a point of the set (−10^4 ≤ X, Y ≤ 10^4 ). Within each test case, no two points have the same location.
The last test case is followed by a line containing one zero.
Output
For each test case output a single line with a single integer representing the number of points in one of the largest subsets of the input that are cocircular.
Sample
Input 7 -10 0 0 -10 10 0 0 10 -20 10 -10 20 -2 4 4 -10000 10000 10000 10000 10000 -10000 -10000 -9999 3 -1 0 0 0 1 0 0 Output 5 3 2
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[Rampage] Blue.Mary:
2010-12-21 02:33:45
My solution has time complexity O(n^3logn). |
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David Gómez:
2010-12-20 21:49:10
Can O(N^4) get AC? |
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Mahesh Chandra Sharma:
2010-12-15 21:43:36
What is the time limit for C++? |
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Shaka Shadows:
2010-11-08 19:32:58
Problem corrected fellows. :) |
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.:: Pratik ::.:
2010-11-08 19:26:35
Fortune cookie is correct, can this be corrected?
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fortune cookie:
2010-11-08 19:26:35
It seems that the judge input terminates with n=0 or EOF (no zero). |
Added by: | psetter |
Date: | 2010-11-04 |
Time limit: | 1s-1.419s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | ACM ICPC2010 – Latin American Regional |