【题目链接】

• 点击打开链接

【思路要点】

• 定义$f_i$表示$i$个点的连通图的个数。

• 考虑用所有图减去不连通的图的个数，枚举$1$号节点所在联通块大小，有

  $f_i=g_i-\sum_{j=1}^{i-1}\binom{i}{j}*f_j*g_{i-j},\ \ g_i=2^{\binom{i}{2}}$

• 记$dp_{i,j}$表示共有$i$个点，恰好$j$个联通块的图的个数，枚举$1$号节点所在联通块大小，有

  $dp_{i,j}=\sum_{k=1}^{i}\binom{i}{k}*f_k*dp_{i-k,j-1}$

• 时间复杂度$O(N^3)$，需要卡常。

【代码】

#include<bits/stdc++.h>

using namespace std;
const int MAXN = 505;
const int P = 998244353;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
int n, m, fac[MAXN], inv[MAXN];
int f[MAXN], g[MAXN], dp[MAXN][MAXN];
int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0) return 1ll * tmp * tmp % P;
  else return 1ll * tmp * tmp % P * x % P;
}
int getc(int x, int y) {
  return 1ll * fac[x] * inv[y] % P * inv[x - y] % P;
}
int main() {
  read(n), read(m);
  fac[0] = 1;
  for (int i = 1; i <= n; i++)
      fac[i] = 1ll * fac[i - 1] * i % P;
  inv[n] = power(fac[n], P - 2);
  for (int i = n - 1; i >= 0; i--)
      inv[i] = inv[i + 1] * (i + 1ll) % P;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
      g[i] = power(2, i * (i - 1) / 2);
  for (int i = 1; i <= n; i++) {
      f[i] = g[i];
      for (int j = 1; j < i; j++)
          f[i] = (f[i] - 1ll * f[j] * g[i - j] % P * getc(i - 1, j - 1) % P + P) % P;
  }
  for (int i = 1; i <= n; i++)
  for (int j = 1; j <= i && j <= m; j++) {
      int tmp = 0;
      for (int k = 1; k <= i && i - k >= j - 1; k++)
          tmp = (tmp + 1ll * f[k] * dp[i - k][j - 1] % P * getc(i - 1, k - 1) % P) % P;
      dp[i][j] = tmp;
  }
  int ans = dp[n][m];
  if (m == 1) ans = (ans - 1 + P) % P;
  writeln(ans);
  return 0;
}