Submission #441201

#	Time	Username	Problem	Language	Result	Execution time	Memory
441201		sinamhdv	Pipes (CEOI15_pipes)	C++11	90 / 100	1431 ms	24752 KiB

pipes

// CEOI15_pipes
// 2nd solution
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1000 * 1000 * 1000 + 7;
const int INF = 1e9 + 100;
const ll LINF = 1e18 + 100;

#ifdef DEBUG
#define dbg(x) cout << #x << " = " << (x) << endl << flush;
#define dbgr(s, f) { cout << #s << ": "; for (auto _ = (s); _ != (f); _++) cout << *_ << ' '; cout << endl << flush; }
#else
#define dbg(x) ;
#define dbgr(s, f) ;
#endif
#define FOR(i, a, b) for (int i = (a); i < (int)(b); i++)
#define fast_io ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define mp make_pair
#define fr first
#define sc second
#define endl '\n'

#define MAXN 100010

int n, m;
vector<int> adj[MAXN];
bool mark[MAXN];
int val[MAXN], h[MAXN], epar[MAXN];
int ef[2 * MAXN], eto[2 * MAXN];

int tree[MAXN], be[MAXN];
int getroot(int x, int *dpar)
{
	return dpar[x] == x ? x : dpar[x] = getroot(dpar[x], dpar);
}
bool merge(int x, int y, int *dpar)
{
	x = getroot(x, dpar);
	y = getroot(y, dpar);
	if (x == y) return false;
	dpar[x] = y;
	return true;
}

void dfs1(int v)
{
	mark[v] = true;
	for (int e : adj[v])
	{
		int u = ef[e] ^ eto[e] ^ v;
		if (mark[u])
		{
			if (h[u] < h[v] && e != epar[v]) val[u]--, val[v]++;
			continue;
		}
		h[u] = h[v] + 1;
		epar[u] = e;
		dfs1(u);
	}
}

void dfs2(int v)
{
	mark[v] = true;
	for (int e : adj[v])
	{
		int u = ef[e] ^ eto[e] ^ v;
		if (!mark[u]) dfs2(u), val[v] += val[u];
	}
	if (epar[v] && val[v] == 0) cout << ef[epar[v]] << ' ' << eto[epar[v]] << endl;
}

int32_t main(void)
{
	fast_io;
	cin >> n >> m;

	iota(tree, tree + n + 5, 0);
	iota(be, be + n + 5, 0);

	int ecnt = 0;

	FOR(i, 0, m)
	{
		int x, y;
		cin >> x >> y;
		if (merge(x, y, tree) || merge(x, y, be))
		{
			ecnt++;
			ef[ecnt] = x;
			eto[ecnt] = y;
			adj[x].pb(ecnt);
			adj[y].pb(ecnt);
		}
	}

	FOR(i, 1, n + 1) if (!mark[i]) dfs1(i);

	memset(mark, 0, sizeof(mark));
	FOR(i, 1, n + 1) if (!mark[i]) dfs2(i);

	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 0 / 10