HDU - 4082 （几何）

题意：给你n个点然后让你找三个点使得拥有最大的相似三角形

题解：可能有重复的点要去除了，接点判断一下是否能组成三角形 ，最后用向量求角度判断相似

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<stack>
using namespace std;
#define mes(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(i = a; i <= b; i++)
#define dec(i,a,b) for(i = b; i >= a; i--)
#define fi first
#define se second
#define ls rt<<1
#define rs rt<<1|1
#define mid (L+R)/2
#define lson ls,L,mid
#define rson rs,mid+1,R
typedef double db;
typedef long long int ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int mx = 1e5+5;
const int x_move[] = {1,-1,0,0,1,1,-1,-1};
const int y_move[] = {0,0,1,-1,1,-1,1,-1};
int n,m;
struct point{
	ll x,y;
	bool operator<(const point &a)const{
		return x*a.y<a.x*y;
	}
}a[mx],s[3],t[3];
db dis[3];
db dist(point a,point b){
	db x = a.x-b.x;
	db y = a.y-b.y;
	return sqrt(x*x+y*y);
}
point angle(point a,point b,point c){
	db x1 = a.x-b.x;
	db y1 = a.y-b.y;
	db x2 = a.x-c.x;
	db y2 = a.y-c.y;
	a.x = (x1*x2+y1*y2)*(x1*x2+y1*y2);
	a.y = (x1*x1+y1*y1)*(x2*x2+y2*y2);
	return a;
}
bool can(point a,point b,point c,point s[]){
	dis[0] = dist(a,b);
	dis[1] = dist(b,c);
	dis[2] = dist(a,c);
	sort(dis,dis+3);
	if(dis[0]+dis[1]<=dis[2]+1e-6)	return false;
	//if(dis[0]+dis[2]<=dis[1]+1e-6)	return false;
	//if(dis[1]+dis[2]<=dis[0]+1e-6)	return false;
	s[0] = angle(a,b,c);
	s[1] = angle(b,a,c);
	s[2] = angle(c,a,b);
	sort(s,s+3);
	return true;
}
bool judge(point s[],point t[]){
	for(int i = 0; i < 3; i++)
		if(s[i].x*t[i].y!=s[i].y*t[i].x)
			return false;
	return true;
}
map<pair<int,int>,int>mp;
int main(){
	int ca = 1;
	while(scanf("%d",&n)&&n){
		int m = 1;
		mp.clear();
		for(int i = 1; i <= n; i++){
			scanf("%lld%lld",&a[m].x,&a[m].y);
			if(!mp[make_pair(a[m].x,a[m].y)]){
				mp[make_pair(a[m].x,a[m].y)]++;
				m++;
			}
		}
		n = m-1;
		int ans = 0;
		for(int i = 1; i <= n; i++)
			for(int j = i+1; j <= n; j++)
				for(int k = j+1; k <= n; k++)
					if(can(a[i],a[j],a[k],s)){
						int tmp = 0;
						for(int ii = 1; ii <= n; ii++)
							for(int jj = ii+1; jj <= n; jj++)
								for(int kk = jj+1; kk <= n; kk++)
									if(can(a[ii],a[jj],a[kk],t)&&judge(t,s))
										tmp++;
						ans = max(tmp,ans);
					}
		printf("%d\n",ans);
	}
	return 0;
}