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+1using 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);    }}