{HDU}{4082}{Hou Yi's secret}{几何}

11年北京赛区的B题，数据量很小暴力搜索的计算几何，但是有很多小地方需要注意，WA了N次：

• 注意对点判重，否则乎会生成很多一样的三角形

• 三个点在同一条直线上的情况

• 判断相似性的时候不需要判断角度，只需要对边进行比例判定即可
  
  

#include <iostream>

#include <string>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <memory>

#include <cmath>

#include <bitset>

#include <queue>

#include <vector>

#include <stack>

using namespace std;



const int MAXN = 20;



#define CLR(x,y) memset(x,y,sizeof(x))

#define MIN(m,v) (m)<(v)?(m):(v)

#define MAX(m,v) (m)>(v)?(m):(v)

#define ABS(x) ((x)>0?(x):-(x))

#define rep(i,x,y) for(i=x;i<y;++i)



typedef struct{

	int a,b,c;

}Node;

Node node[100000];

int N,tt,ans;

int a[MAXN][2];

bool _check(int a1,int b1,

		int a2,int b2,

		int a3,int b3)

{

	int x1 = a1-a2;

	int y1 = b1-b2;

	int x2 = a2-a3;

	int y2 = b2-b3;

	if ( x1*y2 - x2*y1 == 0)

		return false;

	return true;

}

int _dist(const int & a1, const int& b1,

		const int &a2, const int& b2)

{

	return (a1-a2)*(a1-a2)+(b1-b2)*(b1-b2);

}

bool _sim(const int& i, const int& j)

{

	if( node[i].a*node[j].b - node[j].a*node[i].b != 0)

		return false;

	if( node[i].a*node[j].c - node[j].a*node[i].c != 0)

		return false;

	return true;

}

int work()

{

	int i,j,tmp,k;

	int x,y;

	int v[300][300];

	CLR(v,0);

	int n = 0;

	rep(i,0,N){

		scanf("%d%d",&x,&y);

		if( v[x+150][y+150] )

			continue;

		v[x+150][y+150] = true;

		a[n][0] = x;

		a[n][1] = y;

		++n;

	}

	int cnt = 0;

	int e[3];

	rep(i,0,n)

		rep(j,i+1,n)

			rep(k,j+1,n){

				if(!_check(a[i][0],a[i][1],

							a[j][0],a[j][1],

							a[k][0],a[k][1]))

					continue;



				e[0] = _dist(a[i][0],a[i][1],a[j][0],a[j][1]);

				e[1] = _dist(a[i][0],a[i][1],a[k][0],a[k][1]);

				e[2] = _dist(a[k][0],a[k][1],a[j][0],a[j][1]);

				sort(e,e+3);

				node[cnt].a = e[0];

				node[cnt].b = e[1];

				node[cnt].c = e[2];

				++cnt;

			}

	int mmax;

	mmax = 0;

	if( cnt > 1 ){

		tmp = 1;

		rep(i,0,cnt){

			tmp = 1;

			rep(j,i+1,cnt) {

				if(_sim(i,j))

					++tmp;

			}

			if( mmax < tmp )

				mmax = tmp;

		}

		printf("%d\n",mmax);

	}else

		printf("%d\n",cnt);

	return 0;

}

int main()

{

	while( scanf("%d",&N) , N ){

		work();

	}

	return 0;

}