摸鱼过法，多谢楼上老大哥的解答

2进制状态表达

利用哈希表存储当前已经到达的状态记忆化搜索， 内存什么的就不管了(╯‵□′)╯︵┻━┻(绕过不会的dp想法)

#include <iostream>
#include <unordered_set>
#include <unordered_map>
using namespace std;

const int N = 110, K = 22;
int pa[N];
int n, m, k; 
int ans = -1;

unordered_map<int, int> dps;

void dfs(int tp, int status, int ns, unordered_map<int, unordered_set<int>>& ms)
{
    // cout<<"tp is "<<tp<<endl;

    if(dps.count(status) && dps[status] <= ns)return;
    dps[status] = ns;
    while(tp <= m && ((status >> tp) & 1))++tp;
    if(tp == m + 1)
    {
        // cout<<"status "<<status<<endl;
        if(ans < 0 || ns < ans)ans = ns;
        return;
    }

    {
        // cout<<"else situation!"<<endl;
        for(auto ni:ms[tp])
        {
            // cout<<"ni is "<<ni<<endl;
            dfs(tp+1, status|pa[ni], ns+1, ms);
        }
    }
}

int main()
{
    cin>>n>>m>>k;
    unordered_map<int, unordered_set<int>> ms;
    int cs = 0;
    for(int i = 1; i <= n; ++i)
    {
        int t;
        for(int j = 0; j < k; ++j)
        {
            cin>>t;
            pa[i] = pa[i] | (1<<t);
            ms[t].insert(i);
        }
        cs |= pa[i];

        // cout<<hex<<pa[i]<<endl;
    }
    if(cs != (1 << (m+1)) - 2)cout<<"-1"<<endl;
    else
    {
        dfs(1, 0, 0, ms);
        cout<<ans<<endl;
    }

    return 0;
}