网络流。
原点向每个题目连边，容量为 $1$。
每个题目向对应的类型连边，容量为 $1$。
每个类型向汇点连边，容量为它所需的题目数量。

然后相当于求最小割，转化为最大流，于是做完了。
输出方案直接搜一下就好了。

#include <bits/stdc++.h>
using namespace std;
const int N = 1015, M = 2e4 + 15;
const long long INF = 0x3f3f3f3f;

int h[N], e[M], w[M], ne[M], idx = 0;
void add(int a, int b, int c) {
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
    e[idx] = a, w[idx] = 0, ne[idx] = h[b], h[b] = idx++;
}

int n, m, S, T;
int d[N], now[N];

bool bfs() {
    queue<int> q; q.push(S);
    memset(d, 0, sizeof d);
    d[S] = 1, now[S] = h[S];
    while (q.size()) {
        int u = q.front(); q.pop();
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            if (d[v] == 0 && w[i]) {    //未被访问过且该边有残余容量 
                d[v] = d[u] + 1;
                now[v] = h[v];
                if (v == T) return true;
                q.push(v);
            }
        }
    }
    return false;
}

long long find(int u, long long lim) {
    if (u == T) return lim;
    long long flow = 0;
    for (int i = now[u]; ~i && flow < lim; i = ne[i]) {
        now[u] = i;  // 当前弧优化
        int v = e[i];
        if (d[v] == d[u] + 1 && w[i]) {
            int x = find(v, min(w[i] * 1ll, lim - flow));
            if (!x) d[v] = -1;
            w[i] -= x, w[i ^ 1] += x, flow += x;
        }
    }
    return flow;
}

long long dinic() {
    long long ans = 0, res;
    while (bfs()) {
        while (res = find(S, INF)) ans += res;
    }
    return ans;
}

int sum = 0;
vector<int> ans[N];

int main() {
    memset(h, -1, sizeof h);
    scanf("%d%d", &m, &n);
    S = 0, T = n + m + 1;
    for (int i = 1; i <= n; i++) add(S, i, 1);
    for (int i = 1, need; i <= m; i++) scanf("%d", &need), add(n + i, T, need), sum += need;
    for (int i = 1; i <= n; i++) {
        int k; scanf("%d", &k);
        while (k--) {
            int j; scanf("%d", &j);
            add(i, n + j, 1);
        }
    }
    if (dinic() != sum) return puts("No Solution!"), 0;
    for (int u = 1; u <= n; u++) {
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            if (v > n && w[i] == 0) ans[v - n].push_back(u);
        }
    }
    for (int i = 1; i <= m; i++) {
        printf("%d: ", i);
        for (int j : ans[i]) printf("%d ", j);
        puts("");
    }
    return 0;
}