优先队列bfs + 状态压缩

不难想到，最小团就是空集，我们可以向这个空集里加点，来找到k小团

具体做法就是用优先队列，把权值小的团出队，为了不重复加点，我们可以记录一下最后几个被加进团的点，这样下次直接从该点之后遍历，这样就可以把所有点都遍历一次了。

用bitset来保存点连接状态可以直接判断该点是否与团的每个点相连

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long LL;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)){        w |= ch == '-', ch = getchar();    }    while(isdigit(ch)){        ret = (ret << 3) + (ret << 1) + (ch ^ 48);        ch = getchar();    }    return w ? -ret : ret;}inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }template 
     
      inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 200;int n, k, val[N];struct Node{    LL w;    int s;    bitset<105> clique;    bool operator < (const Node &rhs) const {        return w > rhs.w;    }    Node(){}    Node(LL w, int s, bitset<105> clique): w(w), s(s), clique(clique){}};bitset<105> e[N];LL solve(){    priority_queue
      
        pq;    int cnt = 0;    bitset<105> raw;    raw.reset();    pq.push(Node(0, 0, raw));    while(!pq.empty()){        Node cur = pq.top(); pq.pop();        cnt ++;        if(cnt == k) return cur.w;        for(int i = cur.s + 1; i <= n; i ++){            if((e[i] & cur.clique) == cur.clique){                bitset<105> b(cur.clique);                b.set(i, 1);                pq.push(Node(cur.w + val[i], i, b));            }        }    }    return -1;}int main(){    n = read(), k = read();    for(int i = 1; i <= n; i ++) val[i] = read();    string s;    for(int i = 1; i <= n; i ++){        cin >> s;        for(int j = 1; j <= n; j ++){            e[i].set(j, s[j - 1] - '0');        }    }    printf("%lld\n", solve());    return 0;}