제출 #1166607

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1166607		monkey133	Marshmallow Molecules (CCO19_day2problem2)	C++20	25 / 25	64 ms	10568 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

set<ll> adj[100005];

int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n, m, ans = 0;
	cin >> n >> m;
	for (ll i=0; i<m; ++i)
	{
		ll x, y;
		cin >> x >> y;
		if (x > y) swap(x ,y);
		adj[x].insert(y);
	}
	for (ll i=1; i<=n; ++i)
	{
	  ans += adj[i].size();
		if (adj[i].empty()) continue;
		// all node in adj[i] connect
		ll root = *adj[i].begin();
		adj[i].erase(root);
		if (adj[root].size() < adj[i].size()) swap(adj[i], adj[root]);
		for (ll u: adj[i])  adj[root].insert(u);
		adj[i].clear();
	}
	cout << ans << endl;
	return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 5 / 5
• Subtask 4: 10 / 10