Codeforces732F Tourist Reform

求出无向图的所有边双联通分量，然后缩点就成了一颗树。

然后我们选取最大的那个边双联通分量作为根，这样我们就可以确定所有割边的方向了。

对于边双联通分量里面的边，我们随便dfs一下就可以把它变成强连通分量，方向也就确定了。

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

vector<int> G[400010];
vector<pair<int, int>> G2[400010];
set<pair<int, int>> bridge;
set<pair<int, int>> ansset;
pair<int, int> old[400010];
int vis[400010];
int dfn[400010];
int low[400010];
int tag[400010];

void get_bridge(int cur, int father, int dep)
{
    vis[cur] = 1;
    dfn[cur] = low[cur] = dep;
    int children = 0;
    for (auto to : G[cur])
    {
        if (to != father && vis[to] == 1)
        {
            if (dfn[to] < low[cur])
                low[cur] = dfn[to];
        }
        if (vis[to] == 0)
        {
            get_bridge(to, cur, dep + 1);
            children++;
            if (low[to] < low[cur])
                low[cur] = low[to];
            if (low[to] > dfn[cur])
                bridge.insert({cur, to}), bridge.insert({to, cur});
        }
    }
    vis[cur] = 2;
}

int dfs(int u, int tot)
{
    int cnt = 1;
    vis[u] = true;
    tag[u] = tot;
    for (auto to : G[u])
    {
        if (ansset.find({u, to}) == ansset.end() && ansset.find({to, u}) == ansset.end() && bridge.find({u, to}) == bridge.end())
        {
            ansset.insert({u, to});
            if (!vis[to])
                cnt += dfs(to, tot);
        }
    }
    return cnt;
}

void dfs2(int cur, int fa)
{
    for (auto e : G2[cur])
    {
        int nxt = tag[e.second];
        if (nxt != fa)
        {
            ansset.insert({e.second, e.first});
            dfs2(nxt, cur);
        }
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    int u, v;
    for (int i = 0; i < m; i++)
    {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
        old[i] = {u, v};
    }
    get_bridge(1, -1, 0);
    memset(vis, 0, sizeof(vis));
    int ans = 0;
    int tot = 0;
    int idx = 0;
    for (int i = 1; i <= n; i++)
    {
        if (!vis[i])
        {
            int siz = dfs(i, ++tot);
            if (siz > ans)
                ans = siz, idx = tot;
        }
    }
    for (auto e : bridge)
        G2[tag[e.first]].push_back(e);
    dfs2(idx, -1);
    printf("%d
", ans);
    for (int i = 0; i < m; i++)
    {
        if (ansset.find(old[i]) == ansset.end())
            printf("%d %d
", old[i].second, old[i].first);
        else
            printf("%d %d
", old[i].first, old[i].second);
    }
    return 0;
}