题解区很离谱

void solve(int l, int r, vector<int> &v) {
    if (v.empty()) return;
    if (l == r) {   
        for (int &x : v) E[x].t = l;
        return;
    }
    int mid = (l + r) >> 1;
    vector<int> ll, rr, fl;
    init();
    for (int &x : v) {
        if (E[x].t <= mid) addedge(E[x].u, E[x].v), ll.push_back(x);
        else rr.push_back(x);
    }
    for (int &x : tmp) {
        if (!dfn[x]) tarjan(x);
    }
    
    vector<pair<int, int> > need; // 需要连起来的边
    for (int &x : ll) {
        if (block[E[x].u] == block[E[x].v]) fl.push_back(x), need.push_back({E[x].u, E[x].v});
        else rr.push_back(x);
    }
    solve(l, mid, fl);

    for (auto &x : need) { // 用并查集的根代表这个强连通分量即可
        f[find(x.second)] = find(x.first);
    }
    for (int &x : rr) {
        E[x].u = find(E[x].u);
        E[x].v = find(E[x].v);
    }
    solve(mid + 1, r, rr);
}