Submission #673848

#	Time	Username	Problem	Language	Result	Execution time	Memory
673848		benjaminkleyn	Pipes (CEOI15_pipes)	C++17	100 / 100	1061 ms	14208 KiB

pipes

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int max_n = 100000;

int mem[2 * max_n];
int *par[2];

int find(const int &u, const int &dsu)
{
    return (par[dsu][u] == u ? u : par[dsu][u] = find(par[dsu][u], dsu));
}

bool same_set(const int &u, const int &v, const int &dsu)
{
    return find(u, dsu) == find(v, dsu);
}

void unite(const int &u, const int &v, const int &dsu)
{
    par[dsu][find(v, dsu)] = find(u, dsu);
}


int n, m, timer = 0;
vector<int> g[max_n];
int edge_xor[2 * max_n], edge_cnt = 0;
int *ti, *lo;

void dfs(const int &u, const int &p)
{
    ti[u] = lo[u] = ++timer;
    for (const int &e : g[u])
        if (e != p)
        {
            int v = edge_xor[e] ^ u;
            if (ti[v])
                lo[u] = min(lo[u], ti[v]);
            else
            {
                dfs(v, e);
                lo[u] = min(lo[u], lo[v]);

                if (lo[v] > ti[u])
                    cout << u + 1 << ' ' << v + 1 << '\n';
            }
        }
}


int main()
{
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> m;

    par[0] = &mem[0];
    par[1] = &mem[max_n];
    for (int i = 0; i < n; i++)
        par[0][i] = par[1][i] = i;
    
    for (int i = 0, a ,b; i < m; i++)
    {
        cin >> a >> b;
        a--, b--;

        if (!same_set(a, b, 0))
            unite(a, b, 0);
        else if (!same_set(a, b, 1))
            unite(a, b, 1);
        else continue;

        g[a].push_back(edge_cnt);
        g[b].push_back(edge_cnt);
        edge_xor[edge_cnt] = a ^ b;
        edge_cnt++;
    }

    ti = &mem[0];
    lo = &mem[max_n];

    for (int i = 0; i < n; i++)
        ti[i] = lo[i] = 0;

    for (int i = 0; i < n; i++)
        if (!ti[i])
            dfs(i, -1);

    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: 10 / 10