Enumeration. The enumeration point complexity is n ^ 3.
You can also enumerate edge N * n * log (n ).
Poj 1118 must judge that 0 exits.
#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); }}
HDU 1432 lining up (poj 1118)