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

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

 int dfs(int u, int tot)
 {
     int cnt = ;
     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 = ; i < m; i++)
     {
         scanf("%d%d", &u, &v);
         G[u].push_back(v);
         G[v].push_back(u);
         old[i] = {u, v};
     }
     get_bridge(, -, );
     memset(vis, , sizeof(vis));
     int ans = ;
     int tot = ;
     int idx = ;
     for (int i = ; 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, -);
     printf("%d\n", ans);
     for (int i = ; i < m; i++)
     {
         if (ansset.find(old[i]) == ansset.end())
             printf("%d %d\n", old[i].second, old[i].first);
         else
             printf("%d %d\n", old[i].first, old[i].second);
     }
     return ;
 }