Codeforces Round #533 (Div. 2) E - Helping Hiasat 最大团

E - Helping Hiasat

裸的最大团，写了一种 2 ^ (m / 2)  * (m / 2)的复杂度的壮压， 应该还有更好的方法。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 2e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = ;
const double eps = 1e-;

int n, m, op;
LL e[];
int a[<<], b[<<];
string name;
set<int> Set;
map<string, int> Map;

int main() {
    cin >> n >> m;
    for(int i = ; i <= n; i++) {
        cin >> op;
        if(op == ) {
            Set.clear();
        } else {
            cin >> name;
            if(Map.find(name) == Map.end())
                Map[name] = SZ(Map);
            int x = Map[name];
            for(auto& y : Set) {
                e[x - ] |= 1ll << (y - );
                e[y - ] |= 1ll << (x - );
            }
            Set.insert(x);
        }
    }
    int c1 = m / , c2 = m - c1;
    for(int S = ; S < ( << c1); S++) {
        int p = __builtin_ctz(S), tmp = ;
        a[S] = max(a[S], a[S ^ ( << p)]);
        for(int i = ; i < c1; i++)
            if(!(e[p]>>i&)) tmp |=  << i;
        a[S] = max(a[S], a[tmp & (S ^ ( << p))] + );
    }
    for(int S = ; S < ( << c2); S++) {
        int p = __builtin_ctz(S), tmp = ;
        b[S] = max(b[S], b[S ^ ( << p)]);
        for(int i = ; i < c2; i++)
            if(!(e[p + c1]>>(i+c1)&)) tmp |=  << i;
        b[S] = max(b[S], b[tmp & (S ^ ( << p))] + );
    }
    int ans = ;
    for(int S = ; S < ( << c2); S++) {
        LL tmp = ;
        for(int i = ; i < c2; i++)
            if(S >> i & ) tmp |= e[i + c1];
        for(int i = ; i < c1; i++) tmp ^=  << i;
        tmp &= ( << c1) - ;
        ans = max(ans, b[S] + a[tmp]);
    }
    printf("%d\n", ans);
    return ;
}

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