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답안 #874941

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
874941 2023-11-18T06:59:28 Z Bahamin Pipes (CEOI15_pipes) C++17
50 / 100
 801 ms 22448 KB 
#include <bits/stdc++.h>

using namespace std;

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#ifdef LOCAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif

#define ar array
#define ll long long
#define ld long double
#define sze(x) ((ll)x.size())
#define all(a) (a).begin(), (a).end()
#define sui cout.tie(NULL); cin.tie(NULL); ios_base::sync_with_stdio(false)
#define mset(a, x) memset(a, x, sizeof(a))
typedef priority_queue <ll, vector<ll>, greater<ll> > max_heap;
typedef priority_queue <ll> min_heap;
const ll MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;

ll par[2][MAX_N];
vector<pair<ll, ll>> adj[MAX_N];

ll getpar(ll u, ll k)
{
    return (par[k][u] == u ? u : par[k][u] = getpar(par[k][u], k));
}

bool merge(ll u, ll v, ll k)
{   
    u = getpar(u, k), v = getpar(v, k);
    if (u == v) return false;
    par[k][v] = u;
    return true;
}

ll dp[MAX_N];
ll he[MAX_N];

void dfs(ll u, ll h, ll p, ll p2)
{
    he[u] = h;
    dp[u] = INF;
    for (pair<ll, ll> v : adj[u])
    {
        if (v.second == p) continue;
        if (he[v.first]==-1) dfs(v.first, h + 1, v.second, u), dp[u] = min(dp[v.first], dp[u]);
        else dp[u] = min(dp[u], min(he[v.first], h));
    }
    if (dp[u] >= h && p2 != -1) cout << u << " " << p2 << endl;
}

void solve() {
    ll n, m;
    cin >> n >> m;
    for (ll i = 1; i <= n; i++)
    {
        par[0][i] = par[1][i] = i;
        dp[i] = INF;
        he[i] = -1;
        adj[i].clear();
    }
    ll cnt = 0;
    for (ll i = 0; i < m; i++)
    {
        ll a, b;
        cin >> a >> b;
        if (merge(a, b, 0)) cnt++, adj[a].push_back(make_pair(b, cnt)), adj[b].push_back(make_pair(a, cnt));
        else if (merge(a, b, 1)) cnt++, adj[a].push_back(make_pair(b, cnt)), adj[b].push_back(make_pair(a, cnt));
    }
    for (ll i = 1; i <= n; i++)
    {
        if (he[i]==-1) dfs(i, 0, -1, -1);
    }
}

int main() {
    sui;
    ll tc = 1;
    // cin >> tc;
    for (ll t = 1; t <= tc; t++) {
        solve();
    }
}

Subtask #1
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 4444 KB Output is correct
2 Correct 1 ms 4444 KB Output is correct

Subtask #2
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 4 ms 5212 KB Output is correct
2 Correct 4 ms 4956 KB Output is correct

Subtask #3
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 63 ms 4960 KB Output is correct
2 Correct 62 ms 4952 KB Output is correct

Subtask #4
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 112 ms 6184 KB Output is correct
2 Correct 125 ms 5652 KB Output is correct

Subtask #5
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 186 ms 8836 KB Output is correct
2 Correct 173 ms 8280 KB Output is correct

Subtask #6
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Runtime error 259 ms 17024 KB Memory limit exceeded
2 Halted 0 ms 0 KB -

Subtask #7
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Runtime error 433 ms 19240 KB Memory limit exceeded
2 Halted 0 ms 0 KB -

Subtask #8
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Runtime error 529 ms 22448 KB Memory limit exceeded
2 Halted 0 ms 0 KB -

Subtask #9
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Runtime error 688 ms 22444 KB Memory limit exceeded
2 Halted 0 ms 0 KB -

Subtask #10
0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Runtime error 801 ms 21356 KB Memory limit exceeded
2 Halted 0 ms 0 KB -