답안 #682919

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
682919 2023-01-17T08:56:44 Z ghostwriter Pipes (CEOI15_pipes) C++17
20 / 100
5000 ms 25344 KB
// #pragma GCC optimize ("Ofast")
// #pragma GCC target ("avx2")
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
    #include <debug.h>
#else
    #define debug(...)
#endif
#define ft front
#define bk back
#define st first
#define nd second
#define ins insert
#define ers erase
#define pb push_back
#define pf push_front
#define _pb pop_back
#define _pf pop_front
#define lb lower_bound
#define ub upper_bound
#define mtp make_tuple
#define bg begin
#define ed end
#define all(x) (x).bg(), (x).ed()
#define sz(x) (int)(x).size()
// #define int long long
typedef long long ll; typedef unsigned long long ull;
typedef double db; typedef long double ldb;
typedef pair<int, int> pi; typedef pair<ll, ll> pll;
typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll;
typedef string str;
#define FOR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FOS(i, r, l) for (int i = (r); i >= (l); --i)
#define FRN(i, n) for (int i = 0; i < (n); ++i)
#define FSN(i, n) for (int i = (n) - 1; i >= 0; --i)
#define EACH(i, x) for (auto &i : (x))
#define WHILE while
template<typename T> T gcd(T a, T b) { T d2 = (a | b) & -(a | b); a /= d2; b /= d2; WHILE(b) { a = a % b; swap(a, b); } return a * d2; }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
void _assert(bool statement) { if (statement) return; cerr << "\n>> Assertion failed!\n"; exit(0); }
void _assert(bool statement, const str &message) { if (statement) return; cerr << "\n>> Assertion failed: " << message << '\n'; exit(0); }
void _error(const str &message) { cerr << "\n>> Error: " << message << '\n'; exit(0); }
#define file "TEST"
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); }
/*
----------------------------------------------------------------
    END OF TEMPLATE
----------------------------------------------------------------
    Tran The Bao - ghostwriter
    Training for VOI23 gold medal
----------------------------------------------------------------
    GOAT
----------------------------------------------------------------
*/
int n, m;
struct OnlineBridge {
	#define matrix vector<vi>
	int n;
	vi h, s; // heights, size of the component
	matrix adj, p; // adjacent matrix, sparse table parent
	vpi e, bridge; // edges, vector that contains all the bridges
	map<pi, bool> c; // markup map to mark all edge that's not bridge
	OnlineBridge(int n) : n(n) { // 1 indexed
		h.resize(n + 1, 0);
		s.resize(n + 1, 0);
		adj.resize(n + 1);
		p.resize(n + 1);
		FRN(i, n + 1) p[i].resize(17, 0);
		FOR(i, 1, n) s[i] = 1;
	}
	int getp(int u) { // get parent of current components
		FSN(j, 17)
			if (p[u][j])
				u = p[u][j];
		return u;
	}
	void dfs(int u, int pn) { // reorder tree when a new edge is added (update all needed stat)
		FOR(i, 1, 16) p[u][i] = p[p[u][i - 1]][i - 1];
		EACH(j, adj[u]) {
			int v = (e[j].st == u? e[j].nd : e[j].st);
			if (v == pn) continue;
			h[v] = h[u] + 1;
			p[v][0] = u;
			dfs(v, u);
		}
	}
	int lca(int x, int y) { // find lowest common ancestor
		if (h[x] > h[y]) swap(x, y);
		int diff = h[y] - h[x];
		FRN(i, 17)
			if (diff & (1 << i))
				y = p[y][i];
		if (x == y) return x;
		FSN(i, 17)
			if (p[x][i] != p[y][i]) {
				x = p[x][i];
				y = p[y][i];
			}
		return p[x][0];
	}
	bool mark(int x, int y) { // mark edge (x, y), if (x, y) is already marked then return 0 else 1
		if (x > y) swap(x, y);
		if (c.count({x, y})) return 0;
		c[{x, y}] = 1;
		return 1;
	}
	void add(int x, int y) { // process new edge added to graph
		int px = getp(x);
		int py = getp(y);
		if (px == py) {
			int LCA = lca(x, y);
			WHILE(y != LCA) {
				int pn = p[y][0];
				if (!mark(pn, y)) break;
				y = pn;
			}
			WHILE(x != LCA) {
				int pn = p[x][0];
				if (!mark(pn, x)) break;
				x = pn;
			}
		}
		else {
			if (s[px] < s[py]) {
				swap(x, y);
				swap(px, py);
			}
			adj[x].pb(sz(e));
			adj[y].pb(sz(e));
			e.pb({x, y});
			h[y] = h[x] + 1;
			p[y][0] = x;
			dfs(y, x);
			s[px] += s[py];
		}
	}
	vpi getBridge() {
		bridge.clear();
		EACH(i, e) {
			int u = i.st, v = i.nd;
			if (u > v) swap(u, v);
			if (!c.count({u, v})) bridge.pb({u, v});
		}
		return bridge;
	}
};
void reinit(int test_id) {

}
void input(int test_id) {
	cin >> n >> m;
}
void solve(int test_id) {
	OnlineBridge ob(n);
	WHILE(m--) {
		int u, v;
		cin >> u >> v;
		ob.add(u, v);
	}
	vpi ans = ob.getBridge();
	EACH(i, ans) cout << i.st << ' ' << i.nd << '\n';
}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    // freopen(file".inp", "r", stdin);
    // freopen(file".out", "w", stdout);
    int test_num = 1;
    // cin >> test_num; // comment if the problem does not requires multitest
    FOR(i, 1, test_num) {
        input(i); // input in noninteractive sections for case #i
        solve(i); // main function to solve case #i
        reinit(i); // reinit global data to default used in case #i
    }
    #ifdef LOCAL
        cerr << "\nTime: " << setprecision(5) << fixed << (ldb)clock() / CLOCKS_PER_SEC << "ms.\n";
    #endif
    return 0;
}
/*
----------------------------------------------------------------
From Benq:
    stuff you should look for
        * int overflow, array bounds
        * special cases (n=1?)
        * do smth instead of nothing and stay organized
        * WRITE STUFF DOWN
        * DON'T GET STUCK ON ONE APPROACH
----------------------------------------------------------------
*/
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 340 KB Output is correct
2 Incorrect 0 ms 340 KB Wrong number of edges
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 10 ms 1364 KB Wrong number of edges
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 391 ms 1140 KB Output is correct
2 Correct 348 ms 1108 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 767 ms 2588 KB Output is correct
2 Correct 916 ms 2524 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 1335 ms 5864 KB Output is correct
2 Runtime error 1074 ms 21080 KB Memory limit exceeded
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Runtime error 3177 ms 16644 KB Memory limit exceeded
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Execution timed out 5005 ms 20440 KB Time limit exceeded
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Execution timed out 5083 ms 25056 KB Time limit exceeded
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Execution timed out 5079 ms 25144 KB Time limit exceeded
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Execution timed out 5091 ms 25344 KB Time limit exceeded
2 Halted 0 ms 0 KB -