本题可以用最大流也可以用最大匹配(本质一样)，用dinic最大流好建图，但码量大，匈牙利码量小，建图费点劲。
最大流：依旧是设一个源点一个汇点，对于每一个种类，连一条到汇点的边，capacity为需要的量，对于每一个试题，从源点连一条capacity为1的边到他，从他对每一个其所属的编号种类连一条capacity为1的边，求最大流即可，再找出最小割即可
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL; const int maxm = 1e5+;
const int maxn = 1e3+;
const int INF = 0x3f3f3f3f; struct edge{
int u, v, cap, flow, nex;
} edges[maxm]; int head[maxm], cur[maxm], cnt, level[maxn], capacity[];
vector<int> ans[]; void init() {
memset(head, -, sizeof(head));
} void addedge(int u, int v, int cap) {
edges[cnt] = edge{u, v, cap, , head[u]};
head[u] = cnt++;
} void bfs(int s) {
memset(level, -, sizeof(level));
queue<int> q;
level[s] = ;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
for(int i = head[u]; i != -; i = edges[i].nex) {
edge& now = edges[i];
if(now.cap > now.flow && level[now.v] < ) {
level[now.v] = level[u] + ;
q.push(now.v);
}
}
}
} int dfs(int u, int t, int f) {
if(u == t) return f;
for(int& i = cur[u]; i != -; i = edges[i].nex) {
edge& now = edges[i];
if(now.cap > now.flow && level[u] < level[now.v]) {
int d = dfs(now.v, t, min(f, now.cap - now.flow));
if(d > ) {
now.flow += d;
edges[i^].flow -= d;
return d;
} }
}
return ;
} int dinic(int s, int t) {
int maxflow = ;
for(;;) {
bfs(s);
if(level[t] < ) break;
memcpy(cur, head, sizeof(head));
int f;
while((f = dfs(s, t, INF)) > )
maxflow += f;
}
return maxflow;
} void run_case() {
int k, n, p, u, sum = ;
init();
cin >> k >> n;
int s = , t = n+k+;
for(int i = ; i <= k; ++i) {
cin >> capacity[i];
sum += capacity[i];
addedge(n+i, t, capacity[i]), addedge(t, n+i, );
}
for(int i = ; i <= n; ++i) {
cin >> p;
addedge(s, i, ), addedge(i, s, );
while(p--) {
cin >> u;
addedge(i, u+n, ), addedge(u+n, i, );
}
}
if(dinic(s, t) != sum) {cout << "No Solution!"; return;}
for(int i = ; i <= n; ++i) {
for(int j = head[i]; j != -; j = edges[j].nex) {
if(edges[j].flow) {
ans[edges[j].v - n].push_back(i);
break;
}
}
}
for(int i = ; i <= k; ++i) {
cout << i << ":";
for(auto j : ans[i])
cout << " " << j;
cout << "\n";
}
} int main() {
ios::sync_with_stdio(false), cin.tie();
run_case();
//cout.flush();
return ;
}
Dinic
匈牙利：对于每一个种类都连一条边到其所属的试卷，求最大匹配即可(无代码)