結果
| 問題 | No.1983 [Cherry 4th Tune C] 南の島のマーメイド |
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2022-09-02 14:47:36 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.89.0) |
| 結果 |
AC
|
| 実行時間 | 360 ms / 4,000 ms |
| コード長 | 7,466 bytes |
| 記録 | |
| コンパイル時間 | 2,649 ms |
| コンパイル使用メモリ | 215,600 KB |
| 最終ジャッジ日時 | 2025-02-07 00:36:06 |
|
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 41 |
ソースコード
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,fma,abm,mmx,avx,avx2")
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
#define yn(joken) cout<<((joken) ? "Yes" : "No")<<"\n"
#define YN(joken) cout<<((joken) ? "YES" : "NO")<<"\n"
using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vd = vector<double>;
using vld = vector<long double>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvs = vector<vector<string>>;
using vvc = vector<vector<char>>;
using vvd = vector<vector<double>>;
using vvld = vector<vector<long double>>;
using vvvi = vector<vector<vector<int>>>;
using vvvl = vector<vector<vector<ll>>>;
using vvvvi = vector<vector<vector<vector<int>>>>;
using vvvvl = vector<vector<vector<vector<ll>>>>;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
const int INF = 1e9;
const ll LINF = 2e18;
template <class T>
bool chmax(T& a, const T& b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
if (b < a) {
a = b;
return 1;
}
return 0;
}
bool ispow2(int i) { return i && (i & -i) == i; }
bool ispow2(ll i) { return i && (i & -i) == i; }
template <class T>
vector<T> make_vec(size_t a) {
return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
for (int i = 0; i < int(v.size()); i++) {
is >> v[i];
}
return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
for (int i = 0; i < int(v.size()); i++) {
os << v[i];
if (i < int(v.size()) - 1) os << ' ';
}
return os;
}
static uint32_t RandXor(){
static uint32_t x=123456789;
static uint32_t y=362436069;
static uint32_t z=521288629;
static uint32_t w=88675123;
uint32_t t;
t=x^(x<<11);
x=y; y=z; z=w;
return w=(w^(w>>19))^(t^(t>>8));
}
static long double Rand01(){
return (RandXor()+0.5)*(1.0/UINT_MAX);
}
template <typename T = int>
struct Edge{
int from, to;
T cost;
int idx;
Edge() = default;
Edge(int from, int to, T cost = 1, int idx = -1) : from(from), to(to), cost(cost), idx(idx) {}
operator int() const { return to; }
};
template <typename T = int>
struct Graph{
vector<vector<Edge<T>>> g;
int es;
Graph() = default;
explicit Graph(int n) : g(n), es(0) {}
size_t size() const{
return g.size();
}
void add_directed_edge(int from, int to, T cost = 1){
g[from].emplace_back(from, to, cost, es++);
}
void add_edge(int from, int to, T cost = 1){
g[from].emplace_back(from, to, cost, es);
g[to].emplace_back(to, from, cost, es++);
}
void read(int M, int padding = -1, bool weighted = false, bool directed = false){
for (int i = 0; i < M; i++){
int a, b;
cin >> a >> b;
a += padding;
b += padding;
T c = T(1);
if (weighted) cin >> c;
if (directed) add_directed_edge(a, b, c);
else add_edge(a, b, c);
}
}
inline vector<Edge<T>> &operator[](const int &k){
return g[k];
}
inline const vector<Edge<T>> &operator[](const int &k) const{
return g[k];
}
};
template <typename T = int>
using Edges = vector<Edge<T>>;
template<typename T>
vector<int> bridge_tree_decomposition(Graph<T> &G){
int N=(int)G.g.size();
vector<bool> visited(N);
vector<int> ord(N),low(N),cmp(N,-1);
int ts=0,idx=0;
auto dfs=[&](auto &&self,int v,int p=-1)->void{
visited[v]=true;
ord[v]=ts;
low[v]=ord[v];
ts++;
bool flg=false;
for(auto nv:G[v]){
if(!visited[nv]){
self(self,nv,v);
low[v]=min(low[v],low[nv]);
}
else if(nv!=p){
low[v]=min(low[v],ord[nv]);
}
else{
if(!flg) flg=true;
else low[v]=min(low[v],ord[p]);
}
}
};
auto dfs2=[&](auto &&self,int v)->void{
for(auto nv:G[v]){
if(cmp[nv]!=-1) continue;
if(low[nv]<=ord[v]) cmp[nv]=cmp[v];
else{
cmp[nv]=idx;
idx++;
}
self(self,nv);
}
};
for(int i=0;i<N;i++) if(!visited[i]) dfs(dfs,i);
for(int i=0;i<N;i++){
if(cmp[i]==-1){
cmp[i]=idx;
idx++;
dfs2(dfs2,i);
}
}
return cmp;
}
// merge(x,y):mergeする,未併合ならtrueが,併合済みならfalseが返ってくる
// leader(x):xの根を返す
// size(x):xの属する集合のサイズを返す
// same(x,y):x,yが同じ集合に属するかどうか
// groups():各集合に含まれる要素を返す
struct dsu{
public:
dsu() : _n(0) {}
dsu(int n) : _n(n), parent_or_size(n, -1) {}
int merge(int a, int b){
assert(0 <= a && a < _n);
assert(0 <= b && b < _n);
int x = leader(a), y = leader(b);
if (x == y)
return x;
if (-parent_or_size[x] < -parent_or_size[y])
swap(x, y);
parent_or_size[x] += parent_or_size[y];
parent_or_size[y] = x;
return x;
}
bool same(int a, int b){
assert(0 <= a && a < _n);
assert(0 <= b && b < _n);
return leader(a) == leader(b);
}
int leader(int a){
assert(0 <= a && a < _n);
if (parent_or_size[a] < 0)
return a;
return parent_or_size[a] = leader(parent_or_size[a]);
}
int size(int a){
assert(0 <= a && a < _n);
return -parent_or_size[leader(a)];
}
vector<vector<int>> groups(){
vector<int> leader_buf(_n), group_size(_n);
for (int i = 0; i < _n; i++){
leader_buf[i] = leader(i);
group_size[leader_buf[i]]++;
}
vector<vector<int>> result(_n);
for (int i = 0; i < _n; i++){
result[i].reserve(group_size[i]);
}
for (int i = 0; i < _n; i++){
result[leader_buf[i]].push_back(i);
}
result.erase(
remove_if(result.begin(), result.end(),
[&](const vector<int> &v)
{ return v.empty(); }),
result.end());
return result;
}
private:
int _n;
// root node: -1 * component size
// otherwise: parent
vector<int> parent_or_size;
};
void solve(){
int N,M,Q;
cin>>N>>M>>Q;
vi A(M),B(M);
Graph<int> G(N);
rep(i,M){
cin>>A[i]>>B[i];
A[i]--; B[i]--;
G.add_edge(A[i],B[i]);
}
auto ret=bridge_tree_decomposition(G);
dsu UF(N);
rep(i,M){
if(ret[A[i]]!=ret[B[i]]){
UF.merge(A[i],B[i]);
}
}
while(Q--){
int x,y;
cin>>x>>y;
x--; y--;
yn(UF.same(x,y));
}
}
int main(){
cin.tie(nullptr);
ios::sync_with_stdio(false);
solve();
}