[Codeforces] Round #738 (Div. 2) D1. Mocha and Diana (Easy Version)

Codeforces Round #738 (Div. 2) D1. Mocha and Diana (Easy Version)

Solution

유니언 파인드

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 #include <bits/stdc++.h>

using namespace std;

//Speed
#define Code ios_base::sync_with_stdio(false);
#define By cin.tie(NULL);
#define Sumfi cout.tie(NULL);

//Aliases
using ll= long long;
using lld= long double;
using ull= unsigned long long;

//Constants
const lld pi= 3.141592653589793238;
const ll INF= 1e18;
const ll mod=1e9+7;

//TypeDEf
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef unordered_map<ll,ll> umll;
typedef map<ll,ll> mll;

// Macros
#define ff first
#define ss second
#define pb push_back
#define ist insert
#define has count
#define mp make_pair
#define fl(i,n) for(int i=0;i<n;i++)
#define rl(i,m,n) for(int i=n;i>=m;i--)
#define py cout<<"YES\n";
#define pm cout<<"-1\n";
#define pn cout<<"NO\n";
#define vr(v) v.begin(),v.end()
#define rv(v) v.end(),v.begin()

//Debug
#ifdef AbhinavAwasthi
#define debug(x) cerr << #x<<" "; cerr<<x<<" "; cerr << endl;
#else
#define debug(x);
#endif

// Operator overloads
template<typename T1, typename T2> // cin >> pair<T1, T2>
istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }
template<typename T> // cin >> vector<T>
istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}
template<typename T1, typename T2> // cout << pair<T1, T2>
ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }
template<typename T> // cout << vector<T>
ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }

// Utility functions
template <typename T>
void print(T &&t) { cout << t << "\n"; }
void printarr(ll arr[], ll n){fl(i,n) cout << arr[i] << " ";cout << "\n";}
template<typename T>
void printvec(vector<T>v){ll n=v.size();fl(i,n)cout<<v[i]<<" ";cout<<"\n";}
template<typename T>
ll sumvec(vector<T>v){ll n=v.size();ll s=0;fl(i,n)s+=v[i];return s;}

// Mathematical functions
ll gcd(ll a, ll b){if (b == 0)return a;return gcd(b, a % b);}
ll lcm(ll a, ll b){return (a/gcd(a,b)*b);}
ll moduloMultiplication(ll a,ll b,ll mod){ll res = 0;a %= mod;while (b){if (b & 1)res = (res + a) % mod;b >>= 1;}return res;}
ll powermod(ll x, ll y, ll p){ll res = 1;x = x % p;if (x == 0) return 0;while (y > 0){if (y & 1)res = (res*x) % p;y = y>>1;x = (x*x) % p;}return res;}
ll powerll(ll x, ll n){ll res=1;while(n){if (n & 1){res=res * x;}n>>=1;x*=x;}return res;}


//Graph-dfs
// bool gone[MN];
// vector<int> adj[MN];
// void dfs(int loc){
// gone[loc]=true;
// for(auto x:adj[loc])if(!gone[x])dfs(x);
// }

//To Remember
// vector<array<int, 2>> ranges(n);
// cin >> ranges;

// sort(all(ranges), [] (const array<int, 2>& X, const array<int, 2>& Y){
// return X[1] - X[0] < Y[1] - Y[0];
// });

//Sorting
bool sorta(const pair<int,int> &a,const pair<int,int> &b){return (a.second < b.second);}
bool sortd(const pair<int,int> &a,const pair<int,int> &b){return (a.second > b.second);}

//Bits
string decToBinary(int n){string s="";int i = 0;while (n > 0) {s =to_string(n % 2)+s;n = n / 2;i++;}return s;}
ll binaryToDecimal(string n){string num = n;ll dec_value = 0;int base = 1;int len = num.length();for(int i = len - 1; i >= 0; i--){if (num[i] == '1')dec_value += base;base = base * 2;}return dec_value;}

//Check
bool isPrime(ll n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
bool isPowerOfTwo(int n){if(n==0)return false;return (ceil(log2(n)) == floor(log2(n)));}
bool isPerfectSquare(ll x){if (x >= 0) {ll sr = sqrt(x);return (sr * sr == x);}return false;}

//graph
ll graphParent(vll& v,ll n) {return v[n]==n?n:v[n]=graphParent(v,v[n]);}
void graphMerge(vll& v,ll a,ll b) {ll aa=graphParent(v,a),bb=graphParent(v,b);if(aa>bb){swap(aa,bb);}v[aa]=v[bb]=aa;}

//Code by Sumfi
//Language C++
//Practice->Success

//Code

void sumfi()
{
ll n, x, y, a, b;
cin>>n>>x>>y;
vll m1(n+1), m2(n+1);
vpll res;
fl(i,n+1) m1[i]=m2[i]=i;
fl(i,x) {
cin>>a>>b;
graphMerge(m1,a,b);
}
fl(i,y){
cin>>a>>b;
graphMerge(m2,a,b);
}
for(ll i=1;i<=n;++i){
for(ll j=i+1;j<=n;++j){
ll m1ip=graphParent(m1,i),m1jp=graphParent(m1,j),m2ip=graphParent(m2,i),m2jp=graphParent(m2,j);
if(m1ip!=m1jp and m2ip!=m2jp) {
graphMerge(m1,i,j);
graphMerge(m2,i,j);
res.pb(mp(i,j));
}
}
}
cout<<res.size()<<endl;
for(auto [k,v]:res) cout<<k<<" "<<v<<endl;
}
//Main
int main()
{
Code By Sumfi
ll t;
//cin>>t;
//fl(i,t)
{
sumfi();
}
return 0;
}
//End
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/01/28/PS/Codeforces/div2-738-d1/
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