对时间进行线段树分治，可撤销并查集维护连通性。
把每条边出现的时间看作是几条线段，区间修改。
于是做完了。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 15;

int n, m, q;
pair<int, int> edge[N];
int p[N], sz[N];
int find(int x) { return (p[x] == x) ? x : find(p[x]); }
vector<int> range[N];

struct Tree {
    int l, r;
    vector<int> edge;
} tr[N << 2];

void build(int u, int l, int r) {
    tr[u].l = l, tr[u].r = r;
    if (l == r) return;
    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
}

void change(int u, int l, int r, int k) {
    if (l > r) return;
    if (tr[u].l >= l && tr[u].r <= r) {
        tr[u].edge.push_back(k);
        return;
    }
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) change(u << 1, l, r, k);
    if (r > mid) change(u << 1 | 1, l, r, k);
}

int stk[N], top = 0;
void solve(int u) {
    int pre = top;
    for (int i : tr[u].edge) {
        int x = edge[i].first, y = edge[i].second;
        x = find(x), y = find(y);
        if (x == y) continue;
        if (sz[x] < sz[y]) p[x] = y, sz[y] += sz[x], stk[++top] = x;
        else p[y] = x, sz[x] += sz[y], stk[++top] = y;
    }
    if (tr[u].l == tr[u].r) puts( (sz[find(1)] == n) ? "Connected" : "Disconnected");
    if (tr[u].l != tr[u].r) solve(u << 1), solve(u << 1 | 1);
    while (top > pre) {
        int x = stk[top], y = p[x]; top--;
        sz[y] -= sz[x], p[x] = x;
    }
    if (tr[u].l == tr[u].r) return;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) p[i] = i, sz[i] = 1;
    for (int i = 1; i <= m; i++) range[i].push_back(0);
    for (int i = 1; i <= m; i++) scanf("%d%d", &edge[i].first, &edge[i].second);
    scanf("%d", &q);
    for (int i = 1, c, x; i <= q; i++) {
         scanf("%d", &c);
         while (c--) {
             scanf("%d", &x);
             range[x].push_back(i);
         }
    }
    build(1, 1, q);
    for (int i = 1; i <= m; i++) range[i].push_back(q + 1);
    // for (int i = 1; i <= m; i++, puts(""))
    //     for (int j : range[i]) cout << j << ' ';
    for (int i = 1; i <= m; i++) {
        for (int j = 0; j < range[i].size() - 1; j++) {
            int l = range[i][j] + 1, r = range[i][j + 1] - 1;
            change(1, l, r, i);
            // if (l <= r) cout << '\t' << i << ' ' << l << ' ' << r << endl;
        }
    }
    solve(1);
    return 0;
}