枚举，枚举点 复杂度为n^3。

还能够枚举边的，n*n*log(n)。

POJ 1118 要推断0退出。

#include<cstdio>

#include<cstring>

#include<string>

#include<queue>

#include<algorithm>

#include<map>

#include<stack>

#include<iostream>

#include<list>

#include<set>

#include<vector>

#include<cmath>

#define INF 0x7fffffff

#define eps 1e-8

#define LL long long

#define PI 3.141592654

#define CLR(a,b) memset(a,b,sizeof(a))

#define FOR(i,a,n) for(int i= a;i< n ;i++)

#define FOR0(i,a,b) for(int i=a;i>=b;i--)

#define pb push_back

#define mp make_pair

#define ft first

#define sd second

#define sf scanf

#define pf printf

#define acfun std::ios::sync_with_stdio(false)

#define SIZE 700+1

using namespace std;

struct lx

{

    int x,y;

}p[SIZE];

int n;

int main()

{

    while(~sf("%d",&n))

    //while(~sf("%d",&n),n)

    {

        FOR(i,0,n)

        sf("%d%d",&p[i].x,&p[i].y);

        int ans=0;

        int maxn=0;

        FOR(i,0,n)

        {

            FOR(j,i+1,n)

            {

                maxn=0;

                FOR(k,j+1,n)

                {

                    if((p[j].x-p[i].x)*(p[k].y-p[j].y)==(p[j].y-p[i].y)*(p[k].x-p[j].x))

                        maxn++;

                }

                ans=max(maxn,ans);

            }

        }

        pf("%d\n",ans+2);

    }

}