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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#189997#4423. AMPPZ in the times of diseaseberarchegas#TL 0ms0kbC++2313.7kb2023-09-28 05:01:592023-09-28 05:02:00

Judging History

你现在查看的是最新测评结果

  • [2023-09-28 05:02:00]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2023-09-28 05:01:59]
  • 提交

answer

#include <bits/stdc++.h>
#include <iostream>
// #define int long long
#define ld long double
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = b-1; i>=a ; i--)
#define trav(a, x) for(auto& a : x)
#define allin(a , x) for(auto a : x)
#define Unique(v) sort(all(v)),v.erase(unique(all(v)),v.end());
#define sz(v) ((int)v.size())
using namespace std;
typedef vector<int> vi;
#define y1 abacaba
//#define left oooooopss
#define db(x) cerr << #x <<" == "<<x << endl;
#define db2(x,y) cerr<<#x <<" == "<<x<<", "<<#y<<" == "<<y<<endl;
#define db3(x,y,z) cerr << #x<<" == "<<x<<", "<<#y<<" == "<<y<<", "<<#z<<" == "<<z<<endl;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
// std::mt19937_64 rng64((int) std::chrono::steady_clock::now().time_since_epoch().count());
std::mt19937 rng(
  
//  (int) std::chrono::steady_clock::now().time_since_epoch().count()
   1239010
);
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down
//inline ll mod(ll n, ll m ){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll b,ll e,ll m){
    b%=m;
    ll ans = 1;
    for (; e; b = b * b % m, e /= 2)
        if (e & 1) ans = ans * b % m;
    return ans;
}
// debug:
template<class T>void print_vector(vector<T> &v){
    rep(i,0,sz(v))cout<<v[i]<<" \n"[i==sz(v)-1];
}
void dbg_out() {cerr << endl; }
template<typename Head, typename ... Tail> void dbg_out(Head H,Tail... T){
    cerr << ' ' << H;
    dbg_out(T...);
}
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__), cerr << endl
#else
#define dbg(...) 42
#endif
//

//BEGIN DELAUNAY

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fll;

#define sq(x) ((x)*(ll)(x))

struct pt { // ponto
	int x, y;
	pt(int x_ = 0, int y_ = 0) : x(x_), y(y_) {}
	bool operator < (const pt p) const {
		if (x != p.x) return x < p.x;
		return y < p.y;
	}
	bool operator == (const pt p) const {
		return x == p.x and y == p.y;
	}
	pt operator + (const pt p) const { return pt(x+p.x, y+p.y); }
	pt operator - (const pt p) const { return pt(x-p.x, y-p.y); }
	pt operator * (const int c) const { return pt(x*c, y*c); }
	ll operator * (const pt p) const { return x*(ll)p.x + y*(ll)p.y; }
	ll operator ^ (const pt p) const { return x*(ll)p.y - y*(ll)p.x; }
	friend istream& operator >> (istream& in, pt& p) {
		return in >> p.x >> p.y;
	}
};

struct line { // reta
	pt p, q;
	line() {}
	line(pt p_, pt q_) : p(p_), q(q_) {}
	friend istream& operator >> (istream& in, line& r) {
		return in >> r.p >> r.q;
	}
};

// PONTO & VETOR

ll dist2(pt p, pt q) { // quadrado da distancia
	return sq(p.x - q.x) + sq(p.y - q.y);
}

ll sarea2(pt p, pt q, pt r) { // 2 * area com sinal
	return (q-p)^(r-q);
}

bool col(pt p, pt q, pt r) { // se p, q e r sao colin.
	return sarea2(p, q, r) == 0;
}

bool ccw(pt p, pt q, pt r) { // se p, q, r sao ccw
	return sarea2(p, q, r) > 0;
}

int quad(pt p) { // quadrante de um ponto
	return (p.x<0)^3*(p.y<0);
}

bool compare_angle(pt p, pt q) { // retorna se ang(p) < ang(q)
	if (quad(p) != quad(q)) return quad(p) < quad(q);
	return ccw(q, pt(0, 0), p);
}

pt rotate90(pt p) { // rotaciona 90 graus
	return pt(-p.y, p.x);
}

// RETA

bool isinseg(pt p, line r) { // se p pertence ao seg de r
	pt a = r.p - p, b = r.q - p;
	return (a ^ b) == 0 and (a * b) <= 0;
}

bool interseg(line r, line s) { // se o seg de r intersecta o seg de s
	if (isinseg(r.p, s) or isinseg(r.q, s)
		or isinseg(s.p, r) or isinseg(s.q, r)) return 1;

	return ccw(r.p, r.q, s.p) != ccw(r.p, r.q, s.q) and
			ccw(s.p, s.q, r.p) != ccw(s.p, s.q, r.q);
}

int segpoints(line r) { // numero de pontos inteiros no segmento
	return 1 + __gcd(abs(r.p.x - r.q.x), abs(r.p.y - r.q.y));
}

double get_t(pt v, line r) { // retorna t tal que t*v pertence a reta r
	return (r.p^r.q) / (double) ((r.p-r.q)^v);
}

// POLIGONO

// quadrado da distancia entre os retangulos a e b (lados paralelos aos eixos)
// assume que ta representado (inferior esquerdo, superior direito)
ll dist2_rect(pair<pt, pt> a, pair<pt, pt> b) {
	int hor = 0, vert = 0;
	if (a.second.x < b.first.x) hor = b.first.x - a.second.x;
	else if (b.second.x < a.first.x) hor = a.first.x - b.second.x;
	if (a.second.y < b.first.y) vert = b.first.y - a.second.y;
	else if (b.second.y < a.first.y) vert = a.first.y - b.second.y;
	return sq(hor) + sq(vert);
}

ll polarea2(vector<pt> v) { // 2 * area do poligono
	ll ret = 0;
	for (int i = 0; i < v.size(); i++)
		ret += sarea2(pt(0, 0), v[i], v[(i + 1) % v.size()]);
	return abs(ret);
}

// se o ponto ta dentro do poligono: retorna 0 se ta fora,
// 1 se ta no interior e 2 se ta na borda
int inpol(vector<pt>& v, pt p) { // O(n)
	int qt = 0;
	for (int i = 0; i < v.size(); i++) {
		if (p == v[i]) return 2;
		int j = (i+1)%v.size();
		if (p.y == v[i].y and p.y == v[j].y) {
			if ((v[i]-p)*(v[j]-p) <= 0) return 2;
			continue;
		}
		bool baixo = v[i].y < p.y;
		if (baixo == (v[j].y < p.y)) continue;
		auto t = (p-v[i])^(v[j]-v[i]);
		if (!t) return 2;
		if (baixo == (t > 0)) qt += baixo ? 1 : -1;
	}
	return qt != 0;
}

vector<pt> convex_hull(vector<pt> v) { // convex hull - O(n log(n))
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	if (v.size() <= 1) return v;
	vector<pt> l, u;
	for (int i = 0; i < v.size(); i++) {
		while (l.size() > 1 and !ccw(l.end()[-2], l.end()[-1], v[i]))
			l.pop_back();
		l.push_back(v[i]);
	}
	for (int i = v.size() - 1; i >= 0; i--) {
		while (u.size() > 1 and !ccw(u.end()[-2], u.end()[-1], v[i]))
			u.pop_back();
		u.push_back(v[i]);
	}
	l.pop_back(); u.pop_back();
	for (pt i : u) l.push_back(i);
	return l;
}

ll interior_points(vector<pt> v) { // pontos inteiros dentro de um poligono simples
	ll b = 0;
	for (int i = 0; i < v.size(); i++)
		b += segpoints(line(v[i], v[(i+1)%v.size()])) - 1;
	return (polarea2(v) - b) / 2 + 1;
}

struct convex_pol {
	vector<pt> pol;

	// nao pode ter ponto colinear no convex hull
	convex_pol() {}
	convex_pol(vector<pt> v) : pol(convex_hull(v)) {}

	// se o ponto ta dentro do hull - O(log(n))
	bool is_inside(pt p) {
		if (pol.size() == 0) return false;
		if (pol.size() == 1) return p == pol[0];
		int l = 1, r = pol.size();
		while (l < r) {
			int m = (l+r)/2;
			if (ccw(p, pol[0], pol[m])) l = m+1;
			else r = m;
		}
		if (l == 1) return isinseg(p, line(pol[0], pol[1]));
		if (l == pol.size()) return false;
		return !ccw(p, pol[l], pol[l-1]);
	}
	// ponto extremo em relacao a cmp(p, q) = p mais extremo q
	// (copiado de https://github.com/gustavoM32/caderno-zika)
	int extreme(const function<bool(pt, pt)>& cmp) {
		int n = pol.size();
		auto extr = [&](int i, bool& cur_dir) {
			cur_dir = cmp(pol[(i+1)%n], pol[i]);
			return !cur_dir and !cmp(pol[(i+n-1)%n], pol[i]);
		};
		bool last_dir, cur_dir;
		if (extr(0, last_dir)) return 0;
		int l = 0, r = n;
		while (l+1 < r) {
			int m = (l+r)/2;
			if (extr(m, cur_dir)) return m;
			bool rel_dir = cmp(pol[m], pol[l]);
			if ((!last_dir and cur_dir) or
					(last_dir == cur_dir and rel_dir == cur_dir)) {
				l = m;
				last_dir = cur_dir;
			} else r = m;
		}
		return l;
	}
	int max_dot(pt v) {
		return extreme([&](pt p, pt q) { return p*v > q*v; });
	}
	pair<int, int> tangents(pt p) {
		auto L = [&](pt q, pt r) { return ccw(p, r, q); };
		auto R = [&](pt q, pt r) { return ccw(p, q, r); };
		return {extreme(L), extreme(R)};
	}
};

bool operator <(const line& a, const line& b) { // comparador pra reta
	// assume que as retas tem p < q
	pt v1 = a.q - a.p, v2 = b.q - b.p;
	bool b1 = compare_angle(v1, v2), b2 = compare_angle(v2, v1);
	if (b1 or b2) return b1;
	return ccw(a.p, a.q, b.p); // mesmo angulo
}
bool operator ==(const line& a, const line& b) {
	return !(a < b) and !(b < a);
}

// comparador pro set pra fazer sweep line com segmentos
struct cmp_sweepline {
	bool operator () (const line& a, const line& b) const {
		// assume que os segmentos tem p < q
		if (a.p == b.p) return ccw(a.p, a.q, b.q);
		if (a.p.x != a.q.x and (b.p.x == b.q.x or a.p.x < b.p.x))
			return ccw(a.p, a.q, b.p);
		return ccw(a.p, b.q, b.p);
	}
};

// comparador pro set pra fazer sweep angle com segmentos
pt dir;
struct cmp_sweepangle {
    bool operator () (const line& a, const line& b) const {
        return get_t(dir, a) < get_t(dir, b);
    }
};

typedef struct QuadEdge* Q;
struct QuadEdge {
	int id;
	pt o;
	Q rot, nxt;
	bool used;

	QuadEdge(int id_ = -1, pt o_ = pt(INF, INF)) :
		id(id_), o(o_), rot(nullptr), nxt(nullptr), used(false) {}

	Q rev() const { return rot->rot; }
	Q next() const { return nxt; }
	Q prev() const { return rot->next()->rot; }
	pt dest() const { return rev()->o; }
};

Q edge(pt from, pt to, int id_from, int id_to) {
	Q e1 = new QuadEdge(id_from, from);
	Q e2 = new QuadEdge(id_to, to);
	Q e3 = new QuadEdge;
	Q e4 = new QuadEdge;
	tie(e1->rot, e2->rot, e3->rot, e4->rot) = {e3, e4, e2, e1};
	tie(e1->nxt, e2->nxt, e3->nxt, e4->nxt) = {e1, e2, e4, e3};
	return e1;
}

void splice(Q a, Q b) {
	swap(a->nxt->rot->nxt, b->nxt->rot->nxt);
	swap(a->nxt, b->nxt);
}

void del_edge(Q& e, Q ne) { // delete e and assign e <- ne
	splice(e, e->prev());
	splice(e->rev(), e->rev()->prev());
	delete e->rev()->rot, delete e->rev();
	delete e->rot; delete e;
	e = ne;
}

Q conn(Q a, Q b) {
	Q e = edge(a->dest(), b->o, a->rev()->id, b->id);
	splice(e, a->rev()->prev());
	splice(e->rev(), b);
	return e;
}

bool in_c(pt a, pt b, pt c, pt p) { // p ta na circunf. (a, b, c) ?
	__int128 p2 = p*p, A = a*a - p2, B = b*b - p2, C = c*c - p2;
	return sarea2(p, a, b) * C + sarea2(p, b, c) * A + sarea2(p, c, a) * B > 0;
}

pair<Q, Q> build_tr(vector<pt>& p, int l, int r) {
	if (r-l+1 <= 3) {
		Q a = edge(p[l], p[l+1], l, l+1), b = edge(p[l+1], p[r], l+1, r);
		if (r-l+1 == 2) return {a, a->rev()};
		splice(a->rev(), b);
		ll ar = sarea2(p[l], p[l+1], p[r]);
		Q c = ar ? conn(b, a) : 0;
		if (ar >= 0) return {a, b->rev()};
		return {c->rev(), c};
	}
	int m = (l+r)/2;
	auto [la, ra] = build_tr(p, l, m);
	auto [lb, rb] = build_tr(p, m+1, r);
	while (true) {
		if (ccw(lb->o, ra->o, ra->dest())) ra = ra->rev()->prev();
		else if (ccw(lb->o, ra->o, lb->dest())) lb = lb->rev()->next();
		else break;
	}
	Q b = conn(lb->rev(), ra);
	auto valid = [&](Q e) { return ccw(e->dest(), b->dest(), b->o); };
	if (ra->o == la->o) la = b->rev();
	if (lb->o == rb->o) rb = b;
	while (true) {
		Q L = b->rev()->next();
		if (valid(L)) while (in_c(b->dest(), b->o, L->dest(), L->next()->dest()))
			del_edge(L, L->next());
		Q R = b->prev();
		if (valid(R)) while (in_c(b->dest(), b->o, R->dest(), R->prev()->dest()))
			del_edge(R, R->prev());
		if (!valid(L) and !valid(R)) break;
		if (!valid(L) or (valid(R) and in_c(L->dest(), L->o, R->o, R->dest())))
			b = conn(R, b->rev());
		else b = conn(b->rev(), L->rev());
	}
	return {la, rb};
}

vector<vector<int>> delaunay(vector<pt> v) {
	int n = v.size();
	auto tmp = v;
	vector<int> idx(n);
	iota(idx.begin(), idx.end(), 0);
	sort(idx.begin(), idx.end(), [&](int l, int r) { return v[l] < v[r]; });
	for (int i = 0; i < n; i++) v[i] = tmp[idx[i]];
	assert(unique(v.begin(), v.end()) == v.end());
	vector<vector<int>> g(n);
	bool col = true;
	for (int i = 2; i < n; i++) if (sarea2(v[i], v[i-1], v[i-2])) col = false;
	if (col) {
		for (int i = 1; i < n; i++)
			g[idx[i-1]].push_back(idx[i]), g[idx[i]].push_back(idx[i-1]);
		return g;
	}
	Q e = build_tr(v, 0, n-1).first;
	vector<Q> edg = {e};
	for (int i = 0; i < edg.size(); e = edg[i++]) {
		for (Q at = e; !at->used; at = at->next()) {
			at->used = true;
			g[idx[at->id]].push_back(idx[at->rev()->id]);
			edg.push_back(at->rev());
		}
	}
	return g;
}
// END DELAUNAY

const int N = 1e6 + 10;
int a[N];
int pre[N][2];
int suf[N][2];

struct edg{
    int a, b;
    ll cost;
};

struct dsu{
	vi p , ps;
	dsu(int n){
		p = vi(n+1),ps = vi(n+1,1);
		rep(i,0,n+1) p[i] = i;
	}
	int find(int x){return p[x]==x?x:p[x]=find(p[x]);}
	bool join(int x , int y){
		x = find(x) , y = find(y);
		if(x == y) return 0;
		if(ps[x] > ps[y]) swap(x,y);
		p[x] = y, ps[y] += ps[x];
		return 1;
	}
};	

void solve(){
    int n , k;
    cin >> n >> k;
    map<pair<ll,ll> , int> id;
    vector<pt> v;
    for(int i = 0 ; i < n; i ++){
        pair<ll,ll> a;
        cin >> a.first >> a.second;
        id[a] = i;
        v.push_back(pt(a.first,a.second));
    }
    vector<edg> e;
    auto  p = delaunay(v);
    for(int i = 0 ; i < n; i ++){
        for(auto w : p[i])
            e.push_back({i,w,dist2(v[i] , v[w])});
    }
    sort(all(e) , [&](edg a , edg b){
        return a.cost < b.cost;
    });

    dsu T(n);
    vector<edg> mst;
    for(auto [a,b,c] : e){
        if(T.join(a,b)){
            mst.push_back({a,b,c});
        }
    }
    while(k > 1){
        mst.pop_back();
        k--;
    }
    dsu TT(n);
    map<int,int> group;
    for(auto [a,b,c] : mst){
        TT.join(a,b);
    }
    int xp = 0;
    for(int i = 0 ; i < n; i ++){
        if(!group.count(TT.find(i)))
            group[TT.find(i)] = ++xp;
        cout << group[TT.find(i)] << " ";
    }
    cout << endl;
}

int32_t main(){
    fastio;
    int t;
    cin >> t;
    while(t--){
        solve();
    }

    // math -> gcd it all
    // Did you swap N,M? N=1?
}

详细

Test #1:

score: 0
Time Limit Exceeded

input:

100
100000 20
270505 510725
245104 76414
131477 992924
781607 895592
562469 622976
564354 637966
980036 112090
522903 687218
113871 977855
6615 123673
13847 347618
657794 165707
420561 183995
11367 136391
507836 694877
985069 105115
774110 486921
14319 338715
774937 118145
981468 99089
803866 491315...

output:

1 2 3 4 5 5 6 7 3 8 9 10 11 8 7 6 12 9 13 6 12 10 14 4 15 1 11 16 14 5 2 10 17 15 16 4 13 11 4 17 17 9 2 1 5 14 1 18 12 18 14 18 10 14 14 19 8 1 8 14 14 13 19 9 4 11 13 16 15 13 17 10 16 15 8 10 13 12 19 20 4 6 3 12 3 9 11 15 18 12 2 5 5 14 14 4 12 15 6 14 12 1 8 1 8 14 8 11 13 12 15 17 14 15 7 18 5...

result: