这么简单的题都想了半天，肯定是智商低

看起来对Floyd的掌握还不够熟练呢

总之，把2^k的状态扔在最外围循环就ok了

// luogu-judger-enable-o2#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int MAXN = 50 + 2;int N, M;bool d[MAXN][MAXN][63];int f[MAXN][MAXN];int main(){    cin>>N>>M;memset(f, 0x3f, sizeof(f));    for(int i = 1, u, v; i <= M; i++){        scanf("%d%d", &u, &v);        d[u][v][0] = 1, f[u][v] = 1;    }        for(int p = 1; p < 63; p++)        for(int k = 1; k <= N; k++)            for(int i = 1; i <= N; i++)                for(int j = 1; j <= N; j++)                    if(d[i][k][p - 1] && d[k][j][p - 1])                        d[i][j][p] = 1, f[i][j] = 1;        for(int k = 1; k <= N; k++)        for(int i = 1; i <= N; i++)            for(int j = 1; j <= N; j++)                    f[i][j] = min(f[i][j], f[i][k] + f[k][j]);    cout<
         
          <