poj1118 Lining Up

Lining Up

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 16372		Accepted: 5145

Description

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

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


Your program has to be efficient!

Input

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is 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 that all lie on one line.

Sample Input

5
1 1
2 2
3 3
9 10
10 11
0

Sample Output

3

#include<iostream>
#include<algorithm>
using namespace std;

double k[1002];
double x[1002],y[1002];

int main()
{    
    int n;
    int i,j,p;
    int cnt;
    int num,max_num;
    while(cin>>n)
    {
        for(i=0;i<n;i++)
        {
            cin>>x[i];
            cin>>y[i];
        }
        num=2;
        max_num=2;
        for(i=0;i<n;i++)
        {
            cnt=0;
            for(j=0;j<n;j++)
            {
                if(i==j)
                    continue;
                if(x[i]==x[j])
                    k[cnt]=999999999;
                else
                {
                    k[cnt]=(y[i]-y[j])/(x[i]-x[j]);    
                }
                cnt++;
            }
            sort(k,k+cnt);
            num=2;
            for(p=1;p<cnt;p++)
            {
                if(k[p]==k[p-1])
                {
                    num++;
                    if(max_num<num)
                        max_num=num;
                }
                else
                {
                    num=2;
                    if(max_num<num)
                        max_num=num;
                }
            }
        }
        cout<<max_num<<endl;
    }
    return 0;    
}