Submission #1166589

#	Time	Username	Problem	Language	Result	Execution time	Memory
1166589		monkey133	Marshmallow Molecules (CCO19_day2problem2)	C++20	0 / 25	16 ms	5704 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());

vector<ll> adj[100005];
ll par[100005], siz[100005];
ll find(ll x) {return (par[x]==x)?x:par[x] = find(par[x]);}
void unite(ll x, ll y) {par[find(x)] = find(y);}

int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n, m, ans = 0;
	cin >> n >> m;
	for (ll i=1; i<=n; ++i) par[i] = i;
	for (ll i=0; i<m; ++i)
	{
		ll x, y;
		cin >> x >> y;
		if (x > y) swap(x ,y);
		adj[x].pb(y);
	}
	for (ll i=1; i<=n; ++i) for (ll j = 1; j<adj[i].size(); ++j) unite(adj[i][j-1], adj[i][j]);
	for (ll i=1; i<=n; ++i) siz[find(i)]++; 
	for (ll i=1; i<=n; ++i) { ans += siz[i] * (siz[i]-1)/2;}
	cout << ans + m << endl;
	return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 5
• Subtask 4: 0 / 10