题意：射箭落在n个点，任取三点可构成一个三角形，问最大的相似三角形集（一组互相相似的三角形）的个数。
分析：
1、若n个点中有相同的点，要去重，题目中说射箭会形成洞，任选三个洞构成三角形，因此射在同一点只形成一个洞。
2、二进制枚举子集选出三个点，判断能否构成三角形。
3、因为边长可能为小数，因此用边长的平方判断相似。
(1)将比较的两个三角形三边分别排序
(2)tmpx[0] / tmpy[0] = tmpx[1] / tmpy[1] = tmpx[2] / tmpy[2]，即若满足tmpx[0] * tmpy[1] == tmpy[0] * tmpx[1] && tmpx[1] * tmpy[2] == tmpx[2] * tmpy[1]，则两三角形相似。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const LL MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 1000000 + 10;
using namespace std;
struct Node{
int x, y;
}num[20];
struct tri{
LL a, b, c;
}num1[MAXN];
int cnt;
int cntpoint;
vector<int> v;
LL tmpx[5];
LL tmpy[5];
map<pair<int, int>, int> mp;
bool judge(LL x, LL y){
tmpx[0] = num1[x].a;
tmpx[1] = num1[x].b;
tmpx[2] = num1[x].c;
tmpy[0] = num1[y].a;
tmpy[1] = num1[y].b;
tmpy[2] = num1[y].c;
sort(tmpx, tmpx + 3);
sort(tmpy, tmpy + 3);
if(tmpx[0] * tmpy[1] == tmpy[0] * tmpx[1] && tmpx[1] * tmpy[2] == tmpx[2] * tmpy[1]) return true;
return false;
}
int main(){
int n;
while(scanf("%d", &n) == 1){
if(!n) return 0;
mp.clear();
cntpoint = 0;
int x, y;
for(int i = 0; i < n; ++i){
scanf("%d%d", &x, &y);
if(!mp.count(pair<int, int>(x, y))){
mp[pair<int, int>(x, y)] = 1;
num[cntpoint].x = x;
num[cntpoint++].y = y;
}
}
cnt = 0;
for(int i = 0; i < (1 << cntpoint); ++i){
v.clear();
for(int j = 0; j < cntpoint; ++j){
if(i & (1 << j)){
v.push_back(j);
}
}
if(v.size() == 3){
int tmp11 = (num[v[0]].x - num[v[1]].x) * (num[v[0]].x - num[v[1]].x) +(num[v[0]].y - num[v[1]].y) * (num[v[0]].y - num[v[1]].y);
int tmp22 = (num[v[1]].x - num[v[2]].x) * (num[v[1]].x - num[v[2]].x) +(num[v[1]].y - num[v[2]].y) * (num[v[1]].y - num[v[2]].y);
int tmp33 = (num[v[2]].x - num[v[0]].x) * (num[v[2]].x - num[v[0]].x) +(num[v[2]].y - num[v[0]].y) * (num[v[2]].y - num[v[0]].y);
double tmp1 = sqrt(double(tmp11));
double tmp2 = sqrt(double(tmp22));
double tmp3 = sqrt(double(tmp33));
if(dcmp(tmp1 + tmp2, tmp3) > 0 && dcmp(tmp2 + tmp3, tmp1) > 0 && dcmp(tmp1 + tmp3, tmp2) > 0){
num1[cnt].a = (LL)tmp11;
num1[cnt].b = (LL)tmp22;
num1[cnt].c = (LL)tmp33;
++cnt;
}
}
}
int ans = 0;
for(int i = 0; i < cnt; ++i){
int sum = 1;
for(int j = 0; j < cnt; ++j){
if(i != j){
if(judge(i, j)) ++sum;
}
}
ans = max(ans, sum);
}
printf("%d
", ans);
}
return 0;
}