POJ 1018 & HDU 1432 lining up "simple geometry"

Lining up

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 24786		Accepted: 7767

Description

"How am I ever going to solve this problem?" said the pilot.

Indeed, the pilot is not facing a easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly through the area once in a straight line, and she had to fly over as many points as Pos Sible. All points were given by means of the integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set, all lie on one line. Can you write a program that calculates this number?


Your program have to be efficient!

Input

Input consist several Case,first line of the all case was an integer n (1 < n <), then follow n pairs of Intege Rs. Each pair of integers are separated by one blank and ended by a new-line character. The input ended by n=0.

Output

Output one integer for each input case, representing the largest number of points and all lie on one line.

Sample Input

51 12 23 39 1010 110

Sample Output

3

Source

East Central North America 1994
Original title Link: http://poj.org/problem?id=1118
Enter n points and ask how many points you have on the collinear, direct violence. Three-layer loop. Note: Hangzhou Electric test data is a bit large, cin timeout, but POJ can be too. AC Code:

#include <iostream> #include <cstdio>using namespace std;struct point{    int x, y;} A[705];int Main () {    int n;    Freopen ("Data/1018.txt", "R", stdin);    while (Cin>>n,n)    {        for (int i=0;i<n;i++)            scanf ("%d%d", &a[i].x,&a[i].y);            cin>>a[i].x>>a[i].y;        int maxx=2;        for (int i=0;i<n;i++)        {for            (int j=i+1;j<n;j++)            {                int ans=2;                for (int k=j+1;k<n;k++)                {                    if (a[j].x-a[i].x) * (a[k].y-a[j].y) = = (A[J].Y-A[I].Y) * (a[k].x-a[j].x))                        ans++;                    if (Ans>maxx)                        Maxx=ans;        }}} cout<<maxx<<endl;    }    return 0;}

POJ 1018 & HDU 1432 lining up "simple geometry"