結果

問題 No.1293 2種類の道路
ユーザー Hyado
提出日時 2020-11-20 22:01:02
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 60 ms / 2,000 ms
コード長 3,438 bytes
コンパイル時間 2,190 ms
コンパイル使用メモリ 183,656 KB
実行使用メモリ 14,208 KB
最終ジャッジ日時 2024-07-23 13:00:59
合計ジャッジ時間 3,785 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 2
other AC * 22
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
template<typename T> using V = vector<T>;
template<typename T> using VV = vector<vector<T>>;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define siz(v) (ll)(v).size()
#define rep(i,a,n) for(ll i=a;i<(ll)(n);++i)
#define repr(i,a,n) for(ll i=n-1;(ll)a<=i;--i)
#define ENDL '\n'
typedef pair<int,int> Pi;
typedef pair<ll,ll> PL;
constexpr ll mod = 1000000007; // 998244353;
constexpr ll INF = 1000000099;
constexpr ll LINF = (ll)(1e18 +99);
const ld PI = acos((ld)-1);
const vector<ll> dx={-1,0,1,0},dy={0,1,0,-1};
template<typename T,typename U> inline bool chmin(T& t, const U& u){if(t>u){t=u;return 1;}return 0;}
template<typename T,typename U> inline bool chmax(T& t, const U& u){if(t<u){t=u;return 1;}return 0;}
template<typename T> inline T gcd(T a,T b){return b?gcd(b,a%b):a;}
inline void yes() { cout << "Yes" << ENDL; }
inline void no() { cout << "No" << ENDL; }
template<typename T,typename Y> inline T mpow(T a, Y n) {
T res = 1;
for(;n;n>>=1) {
if (n & 1) res = res * a;
a = a * a;
}
return res;
}
template <typename T> V<T> prefix_sum(const V<T>& v) {
int n = v.size();
V<T> ret(n + 1);
rep(i, 0, n) ret[i + 1] = ret[i] + v[i];
return ret;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
for(auto&& x: vec) is >> x;
return is;
}
template<typename T,typename Y>
ostream& operator<<(ostream& os,const pair<T,Y>& p){
return os<<"{"<<p.fs<<","<<p.sc<<"}";
}
template<typename T> ostream& operator<<(ostream& os,const V<T>& v){
os<<"{";
for(auto e:v)os<<e<<",";
return os<<"}";
}
template<typename ...Args>
void debug(Args&... args){
for(auto const& x:{args...}){
cerr<<x<<' ';
}
cerr<<ENDL;
}
template <typename T> struct UnionFind {
vector<T> Parent;
UnionFind(T N) { Parent = vector<T>(N, -1); }
T root(T A) {
if(Parent[A] < 0) return A;
return Parent[A] = root(Parent[A]);
}
T size(T A) { return -Parent[root(A)]; }
bool connect(T A, T B) {
A = root(A);
B = root(B);
if(A == B) return false;
if(size(A) < size(B)) swap(A, B);
Parent[A] += Parent[B];
Parent[B] = A;
return true;
}
bool same(T A, T B) { return root(A) == root(B); }
};
signed main(){
cin.tie(0);cerr.tie(0);ios::sync_with_stdio(false);
cout<<fixed<<setprecision(20);
ll n,a,b;cin>>n>>a>>b;
UnionFind<ll> au(n),bu(n);
rep(i,0,a){
int x,y;cin>>x>>y;
--x;--y;
au.connect(x,y);
}
rep(i,0,b){
int x,y;cin>>x>>y;
--x;--y;
bu.connect(x,y);
}
ll ans=0;
V<set<int>> rcon(n,set<int>());
rep(i,0,n){
rcon[au.root(i)].emplace(bu.root(i));
}
rep(i,0,n){
int aroot=au.root(i);
if(aroot!=i)continue;
for(auto&& j:rcon[aroot]){
int broot=bu.root(j);
ans+=au.size(aroot)*bu.size(broot);
}
ans-=au.size(aroot);
}
cout<<ans<<ENDL;
}
//! ( . _ . ) !
//CHECK overflow,vector_size,what to output?
//any other simpler approach?
//list all conditions, try mathematical and graphic observation
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