PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4
Setter: Siddhartha Tiwari
Tester: Aryan, Satyam
Editorialist: Devendra Singh
DIFFICULTY:
2522
PREREQUISITES:
Basic geometry
PROBLEM:
You are given a set of N distinct points P_1, P_2, P_3, \ldots, P_N on a 2-D plane.
A triplet (i, j, k) is called a holy triplet if
- 1 \leq i \lt j \lt k \leq N
- P_i, P_j and P_k are non-collinear and
- Any two of the points P_i, P_j and P_k are antipodal points of the circle that passes through all three of them.
Two points on a circle are said to be antipodal points of the circle if they are diametrically opposite to each other.
Find the total number of holy triplets.
EXPLANATION:
An important observation
Consider a holy triplet (P_1,P_2,P_3) such that P_1 and P_3 are antipodal then P_1,P_2,P_3 lie on a semicircle as P_1 and P_3 are diametrically opposite and thus the \angle P_1P_2P_3=\pi / 2. Hence we need to count the number of triplets such that they from a right angled triangle.
The brute force solution is of the time complexity O(n^3) and even implementing a way to check if two of these points are anti-podal is not an easy task.
One way to check whether 3 points P_1,P_2,P_3 form a holy triplet is:
For now, let’s assume these points are not collinear. We
can find out whether the angle between the lines P_1-P_2 and P_1-P_3 is \pi / 2 radians or not. To check this we have to multiply the slopes of these two lines and check if it equal to -1 or not. If it is, then (P_2, P_3) will be antipodal. If it’s not check the same condition for other pairs of lines,
namely (P_2-P_3 and P_2-P_1) and (P_1-P_2 and P_1-P_3). If one of these conditions is true then we have a holy triplet.
Now, this solution can be reduced to O(n^2log(n)). To do that, we have to fix the point that will act as the point other than the two antipodal points in any triplet. Let fixed point be P_1 and we are searching for all pairs (P_2, P_3) in the set of points which satisfy the given requirement. This can be done in O(nlog(n)). Create an array of pairs of integers A[1...n]. A_i
is defined as A_i → (a, b), where a / b is the slope of the line formed by joining the fixed point P_1 and P_i and gcd(a, b) = 1. Now we can say that two points P_i and P_j are antipodal with P_1 as the
third point, if (A[i].a / A[i].b) * (A[j].a / A[j].b) = -1 or A[j].a / A[j].b = -A[i].b / A[i].a. And since
gcd(A[j].a, A[j].b) = gcd(A[i].a, A[i].b) = 1, we can safely say that A[j].a = -A[i].b and A[j].b = A[i].a
OR A[j].a = A[i].b and A[j].b = -A[i].a. Now, we can use a map to store these pairs. Now iterate on the array of points taking P_i as a fixed point and create the array A for each point as shown above. Now iterate for point P_2 and check if there exists a point P_3 antipodal to it. If yes add it to the answer.
TIME COMPLEXITY:
O(N^2log(N)) for each test case.
SOLUTION:
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
class Point {
public:
LL x, y;
Point(LL x, LL y) {
this->x = x;
this->y = y;
}
Point() {
this->x = 0;
this->y = 0;
}
string toString() {
return "{" + to_string(x) + "; " + to_string(y) + "}";
}
};
class Slope {
public:
LL numerator, denominator;
Slope(Point first, Point second) {
numerator = second.y - first.y;
denominator = second.x - first.x;
LL sign = 1, gcd = 1;
if (numerator * denominator < 0)
sign = -1;
numerator = abs(numerator);
denominator = abs(denominator);
if (numerator == 0 || denominator == 0)
numerator *= sign;
else {
gcd = __gcd(numerator, denominator);
numerator /= gcd;
denominator /= gcd;
}
numerator *= sign;
}
Slope(LL numerator, LL denominator) {
this->numerator = numerator;
this->denominator = denominator;
LL sign = 1;
if (this->numerator * this->denominator < 0)
sign = -1;
this->numerator = abs(this->numerator);
this->denominator = abs(this->denominator);
this->numerator *= sign;
}
Slope() {
this->numerator = 0;
this->denominator = 0;
}
Slope getPerpendicular() {
return Slope(-denominator, numerator);
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int T;
cin >> T;
while (T--) {
int N;
cin >> N;
vector<Point> points(N);
for (int i = 0; i < N; i++)
cin >> points[i].x >> points[i].y;
// fix the non-antipodal point in the triplet, and search for antipodal points around it.
int ans = 0;
for (int i = 0; i < N; i++) {
map<pair<LL, LL>, int> count;
vector<Slope> slopes(N);
int zero = 0, infinity = 0;
for (int j = 0; j < N; j++) {
if (i == j) continue;
slopes[j] = Slope(points[i], points[j]);
pair<LL, LL> nPair(slopes[j].numerator, slopes[j].denominator);
if (count.find(nPair) == count.end())
count.insert({nPair, 1});
else
count[nPair]++;
if (slopes[j].numerator == 0) zero++;
if (slopes[j].denominator == 0) infinity++;
}
for (int j = 0; j < N; j++) {
if (i == j) continue;
if (slopes[j].numerator == 0) {
ans += infinity;
continue;
}
else if (slopes[j].denominator == 0) {
ans += zero;
continue;
}
Slope perpendicular = slopes[j].getPerpendicular();
pair<LL, LL> nPair(perpendicular.numerator, perpendicular.denominator);
if (count.find(nPair) != count.end())
ans += count[nPair];
}
}
cout << ans / 2 << "\n";
}
return 0;
}
Tester-1's Solution
/* in the name of Anton */
/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/
#ifdef ARYANC403
#include <header.h>
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif
// y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801
// http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html
template<class Fun> class y_combinator_result {
Fun fun_;
public:
template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {}
template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(*this), std::forward<Args>(args)...); }
};
template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); }
using namespace std;
#define fo(i,n) for(i=0;i<(n);++i)
#define repA(i,j,n) for(i=(j);i<=(n);++i)
#define repD(i,j,n) for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"
typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;
const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}
long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
return readInt(l,r,' ');
}
long long readIntLn(long long l, long long r) {
return readInt(l,r,'\n');
}
string readStringLn(int l, int r) {
return readString(l,r,'\n');
}
string readStringSp(int l, int r) {
return readString(l,r,' ');
}
void readEOF(){
assert(getchar()==EOF);
}
vi readVectorInt(int n,lli l,lli r){
vi a(n);
for(int i=0;i<n-1;++i)
a[i]=readIntSp(l,r);
a[n-1]=readIntLn(l,r);
return a;
}
bool isBinaryString(const string s){
for(auto x:s){
if('0'<=x&&x<='1')
continue;
return false;
}
return true;
}
// #include<atcoder/dsu>
// vector<vi> readTree(const int n){
// vector<vi> e(n);
// atcoder::dsu d(n);
// for(lli i=1;i<n;++i){
// const lli u=readIntSp(1,n)-1;
// const lli v=readIntLn(1,n)-1;
// e[u].pb(v);
// e[v].pb(u);
// d.merge(u,v);
// }
// assert(d.size(0)==n);
// return e;
// }
const lli INF = 0xFFFFFFFFFFFFFFFL;
lli seed;
mt19937 rng(seed=chrono::steady_clock::now().time_since_epoch().count());
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}
class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{ return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y )); }};
void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end()) m.insert({x,cnt});
else jt->Y+=cnt;
}
void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt) m.erase(jt);
else jt->Y-=cnt;
}
bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}
const lli mod = 1000000007L;
// const lli maxN = 1000000007L;
lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .
ii reduce(lli x,lli y){
lli g=__gcd(abs(x),abs(y));
x/=g;y/=g;
if(y<0){
x*=-1;
y*=-1;
}
// if(x==0&&y<0)
// y*=-1;
return {x,y};
}
int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
T=readIntLn(1,10);
lli sumN = 2000;
while(T--)
{
n=readIntLn(3,sumN);
sumN-=n;
vii a(n);
for(auto &x:a){
auto b=readVectorInt(2,-1e9,1e9);
x.X=b[0];
x.Y=b[1];
}
sort(all(a));
(a).erase(unique(all(a)),(a).end());
assert(sz(a)==n);
lli ans=0;
for(int deg90=0;deg90<n;++deg90){
map<ii,lli> count;
lli x=0,y=0;
for(int j=0;j<n;j++){
if(deg90==j)
continue;
auto oth=reduce(-(a[deg90].X-a[j].X),a[deg90].Y-a[j].Y);
if(oth.X==0)
x++;
if(oth.Y==0)
y++;
if(oth.X==0||oth.Y==0)
continue;
ans+=count[oth];
auto cur=reduce(a[deg90].Y-a[j].Y,a[deg90].X-a[j].X);
count[cur]++;
}
ans+=x*y;
}
cout<<ans<<endl;
} aryanc403();
readEOF();
return 0;
}
Tester-2's Solution
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif
#define ll int
/*
------------------------Input Checker----------------------------------
*/
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
ll MAX=1e9;
ll tes_sum=0;
void solve(){
ll n=readIntLn(3,2000);
tes_sum+=n;
vector<ll> x(n+5),y(n+5);
ll ans=0;
set<pair<ll,ll>> check;
for(ll i=1;i<=n;i++){
x[i]=readIntSp(-MAX,MAX); y[i]=readIntLn(-MAX,MAX);
check.insert({x[i],y[i]});
}
assert(check.size()==n);
for(ll i=1;i<=n;i++){
map<pair<ll,ll>,ll> freq;
for(ll j=1;j<=n;j++){
if(i==j){
continue;
}
ll num=y[i]-y[j],denom=x[i]-x[j];
ll g=__gcd(num,denom);
num/=g; denom/=g;
if(denom<0){
denom=-denom,num=-num;
}
freq[{num,denom}]++;
ll l=-denom,r=num;
if(r<0){
r=-r,l=-l;
}
ans+=freq[{l,r}];
}
}
cout<<ans<<"\n";
return;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("error.txt", "w", stderr);
#endif
ll test_cases=readIntLn(1,10);
while(test_cases--){
solve();
}
assert(getchar()==-1);
assert(tes_sum<=2000);
return 0;
}
