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发现自己不会求最大团了可海星

如果将每一个朋友看做点，将两个\(1\)之间存在\(2\)操作的所有朋友之间互相连边，那么我们最后要求的就是这个图的最大独立集。

某个图的最大独立集就是反图的最大团

然后暴力dfs求最大团即可

#include<iostream>
#include<cstdio>
#include<bitset>
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    char c = getchar();
    bool f = 0;
    while(!isdigit(c) && c != EOF){
        if(c == '-')
            f = 1;
        c = getchar();
    }
    if(c == EOF)
        exit(0);
    while(isdigit(c)){
        a = a * 10 + c - 48;
        c = getchar();
    }
    return f ? -a : a;
}

bitset < 41 > Edge[41] , cur , choose;
string mod[41] , s;
int N , M , cnt , maxN;

inline int find(){
    for(int i = 1 ; i <= cnt ; ++i)
        if(mod[i] == s)
            return i;
    mod[++cnt] = s;
    return cnt;
}

void dfs(int x){
    if(choose.count() + x <= maxN)
        return;
    if(!x){
        maxN = choose.count();
        return;
    }
    if((Edge[x] & choose) == choose){
        choose.set(x);
        dfs(x - 1);
        choose.reset(x);
    }
    dfs(x - 1);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    N = read();
    M = read();
    for(int i = 1 ; i <= M ; ++i)
        Edge[i].set();
    for(int i = 1 ; i <= N ; ++i)
        if(read() == 1)
            cur.reset();
        else{
            cin >> s;
            int t = find();
            if(!cur[t]){
                for(int i = 1 ; i <= cnt ; ++i)
                    if(cur[i])
                        Edge[i][t] = Edge[t][i] = 0;
                cur.set(t);
            }
        }
    dfs(M);
    cout << maxN;
    return 0;
}