[CF118E] Bertown roads - Tarjan

Description

有n个交汇点和m条双向道路，一个人可以凭借现有的通道从一个点到达另一个点。确定这样一种方案，使每条道路改为单向行驶，但仍满足能够从任意一个点到达另一任意点的要求

Solution

关键是把图做成若干个环，而无解当且仅当图中有桥

因此我们先用 TARJAN 判断是否有桥，如果没有，构造解法输出，只需要一个不走重复边的 DFS 即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 1e6 + 5;
int n, m;
vector<int> g[N], b[N], inv[N];
int low[N], dfn[N], ind, vis[N];
int flag_bridge;

void make(int p, int q)
{
    inv[p].push_back(g[q].size());
    inv[q].push_back(g[p].size());
    g[p].push_back(q);
    g[q].push_back(p);
    b[p].push_back(0);
    b[q].push_back(0);
}

void tarjan(int p, int from)
{
    low[p] = dfn[p] = ++ind;
    for (int i = 0; i < g[p].size(); i++)
    {
        int q = g[p][i];
        if (q == from)
            continue;
        if (dfn[q] == 0)
        {
            tarjan(q, p);
            low[p] = min(low[p], low[q]);
        }
        else
        {
            low[p] = min(low[p], dfn[q]);
        }
        if (low[q] > dfn[p])
            flag_bridge = 1;
    }
}

void print(int p)
{
    vis[p] = 1;
    for (int i = 0; i < g[p].size(); i++)
    {
        int q = g[p][i];
        if (b[p][i])
            continue;
        b[p][i] = b[q][inv[p][i]] = 1;
        if (!vis[q])
            print(q);
        cout << p << " " << q << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);

    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        make(u, v);
    }
    tarjan(1, 0);
    if (flag_bridge)
    {
        cout << 0 << endl;
    }
    else
    {
        print(1);
    }
}