제출 #210868

#제출 시각아이디문제언어결과실행 시간메모리
210868mode149256게임 (IOI14_game)C++14
42 / 100
1012 ms16476 KiB
/*input

*/
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#include "game.h"
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

const int MX = 1500;

vii liko(MX, vi(MX, 1));
vbb buvo(MX, vb(MX, false));
vi p;
int N;

void initialize(int n) {
	N = n;
	p = vi(n + 1); for (int i = 0; i <= n; i++) p[i] = i;
}

void unionSet(int a, int b) {
	int prad = p[a];
	for (int i = 0; i < N; ++i)
		if (p[i] == prad)
			p[i] = p[b];
}

void merge(int a, int b) {
	unionSet(a, b);
	int tev = p[a];

	vi curr;
	vi kiti;

	for (int i = 0; i < N; ++i)
	{
		if (p[i] != tev) kiti.emplace_back(i);
		else curr.emplace_back(i);
		liko[tev][i] = 0;
		liko[i][tev] = 0;
	}

	for (auto pir : curr) {
		for (auto ant : kiti) {
			liko[tev][p[ant]] += !buvo[pir][ant];
			liko[p[ant]][tev] += !buvo[pir][ant];
		}
	}
}

int hasEdge(int u, int v) {
	buvo[u][v] = buvo[v][u] = true;
	u = p[u];
	v = p[v];

	if (u == v) return 0;

	if (liko[u][v] != 1) {
		liko[u][v]--;
		liko[v][u]--;
		return 0;
	}

	merge(u, v);
	return 1;
}
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