Freezing

Freezing

牛客小白月赛53F

首先有一个朴素的 \(DP\) ， dp[i][j] 表示前 \(i\) 个人中，且最后一个人的状态是 \(j\) 的方案数，然后暴力转移即可，复杂度 \(O(n \times 2^m)\) 无法通过本题。

有一个不能通过本题的优化是，我们可以通过枚举 ~j 的子集，来进行状态转移，赛时可以通过，赛后被出题人卡掉了。代码如下

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i, a, b) for (int i(a); i <= b; ++ i ) 
#define dec(i, a, b) for (int i(b); i >= a; -- i ) 

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); }

const int MOD = 998244353;

inline int mod(int x) {return x >= MOD ? x - MOD : x;}

inline int ksm(int a, int b) {
  int ret = 1; a = mod(a);
  for(; b; b >>= 1, a = 1LL * a * a % MOD) if(b & 1) ret = 1LL * ret * a % MOD;
  return ret;
}

template<int MOD> 
struct modint {
  int x;
  modint() {x = 0; }
  modint(int y) {x = y;}
  inline modint inv() const { return modint{ksm(x, MOD - 2)}; }
  explicit inline operator int() { return x; }
  friend inline modint operator + (const modint &a, const modint& b) { return modint(mod(a.x + b.x)); }
  friend inline modint operator - (const modint &a, const modint& b) { return modint(mod(a.x - b.x + MOD)); }
  friend inline modint operator * (const modint &a, const modint& b) { return modint(1ll * a.x * b.x % MOD); }
  friend inline modint operator / (const modint &a, const modint& b) { return modint(1ll * a.x * b.inv().x % MOD); }
  friend inline modint operator + (const modint &a, const int& b) { return modint(mod(a.x + b)); }
  friend inline modint operator - (const modint &a, const int& b) { return modint(mod(a.x - b + MOD)); }
  friend inline modint operator * (const modint &a, const int& b) { return modint(1ll * a.x * b % MOD); }
  friend inline modint operator / (const modint &a, const int& b) { return modint(1ll * a.x * ksm(b, MOD - 2) % MOD); } 
  friend inline modint operator - (const modint &a) { return modint(mod(MOD - a.x)); }
  friend inline modint& operator += (modint &a, const modint& b) { return a = a + b; }
  friend inline modint& operator -= (modint &a, const modint& b) { return a = a - b; }
  friend inline modint& operator *= (modint &a, const modint& b) { return a = a * b; }
  friend inline modint& operator /= (modint &a, const modint& b) { return a = a / b; }
  friend inline modint& operator += (modint &a, const int& b) { return a = a + b; }
  friend inline modint& operator -= (modint &a, const int& b) { return a = a - b; }
  friend inline modint& operator *= (modint &a, const int& b) { return a = a * b; }
  friend inline modint& operator /= (modint &a, const int& b) { return a = a / b; }
  friend auto &operator >> (istream &i, modint &a) {return i >> a.x; }
  friend auto &operator << (ostream &o, const modint &z) { return o << z.x; }
  inline bool operator == (const modint &b) { return x == b.x; }
  inline bool operator != (const modint &b) { return x != b.x; }
  inline bool operator < (const modint &a) { return x < a.x; }
  inline bool operator <= (const modint &a) { return x <= a.x; }
  inline bool operator > (const modint &a) { return x > a.x; }
  inline bool operator >= (const modint &a) { return x >= a.x; }
  operator int() const {
    return x;
  }
  // inline void
};

typedef modint<MOD> mint;

inline mint ksm(mint a, int b, mint ret = 1) {
	for(; b; b >>= 1, a = a * a ) if(b & 1) ret = ret * a ;
	return ret;
}

constexpr int N = 2E5 + 10;

int n, m;
mint f[1 << 16];// g[1 << 16];
// f[i][j] 表示高八位为 i 且 低八位和j没有交集的方案数

void solve() {
  cin >> n >> m;
  int all = (1 << m) - 1;  
  for (int i = 1; i <= n; i ++ ) {
    string s; cin >> s;
    int x = 0;
    for (int j = 0; j < m; j ++ ) if (s[j] - 'h') {
      x |= 1 << j;
    }
    int y = x ^ all;
    mint now = f[0] + 1;
    for (int j = y; j; j = (j - 1) & y) {
      now += f[j];
    }
    f[x] += now;
  }
  mint ans = 0;
  for (int i = 0; i < (1 << m); i ++ ) ans += f[i];
  cout << ans  << "\n";

}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  solve();
  return 0;
}

下面考虑正确的 \(DP\) ，可以把 \(m\) 位二进制数拆分为两个 \(\frac{m}{2}\) 位的数字。令 dp[i][j] 表示高八位为 \(i\) 且 低八位和 \(j\) 没有交集的方案数，然后进行转移即可，具体过程可以参考代码。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i, a, b) for (int i(a); i <= b; ++ i ) 
#define dec(i, a, b) for (int i(b); i >= a; -- i ) 

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); }

const int MOD = 998244353;

inline int mod(int x) {return x >= MOD ? x - MOD : x;}

inline int ksm(int a, int b) {
  int ret = 1; a = mod(a);
  for(; b; b >>= 1, a = 1LL * a * a % MOD) if(b & 1) ret = 1LL * ret * a % MOD;
  return ret;
}

template<int MOD> 
struct modint {
  int x;
  modint() {x = 0; }
  modint(int y) {x = y;}
  inline modint inv() const { return modint{ksm(x, MOD - 2)}; }
  explicit inline operator int() { return x; }
  friend inline modint operator + (const modint &a, const modint& b) { return modint(mod(a.x + b.x)); }
  friend inline modint operator - (const modint &a, const modint& b) { return modint(mod(a.x - b.x + MOD)); }
  friend inline modint operator * (const modint &a, const modint& b) { return modint(1ll * a.x * b.x % MOD); }
  friend inline modint operator / (const modint &a, const modint& b) { return modint(1ll * a.x * b.inv().x % MOD); }
  friend inline modint operator + (const modint &a, const int& b) { return modint(mod(a.x + b)); }
  friend inline modint operator - (const modint &a, const int& b) { return modint(mod(a.x - b + MOD)); }
  friend inline modint operator * (const modint &a, const int& b) { return modint(1ll * a.x * b % MOD); }
  friend inline modint operator / (const modint &a, const int& b) { return modint(1ll * a.x * ksm(b, MOD - 2) % MOD); } 
  friend inline modint operator - (const modint &a) { return modint(mod(MOD - a.x)); }
  friend inline modint& operator += (modint &a, const modint& b) { return a = a + b; }
  friend inline modint& operator -= (modint &a, const modint& b) { return a = a - b; }
  friend inline modint& operator *= (modint &a, const modint& b) { return a = a * b; }
  friend inline modint& operator /= (modint &a, const modint& b) { return a = a / b; }
  friend inline modint& operator += (modint &a, const int& b) { return a = a + b; }
  friend inline modint& operator -= (modint &a, const int& b) { return a = a - b; }
  friend inline modint& operator *= (modint &a, const int& b) { return a = a * b; }
  friend inline modint& operator /= (modint &a, const int& b) { return a = a / b; }
  friend auto &operator >> (istream &i, modint &a) {return i >> a.x; }
  friend auto &operator << (ostream &o, const modint &z) { return o << z.x; }
  inline bool operator == (const modint &b) { return x == b.x; }
  inline bool operator != (const modint &b) { return x != b.x; }
  inline bool operator < (const modint &a) { return x < a.x; }
  inline bool operator <= (const modint &a) { return x <= a.x; }
  inline bool operator > (const modint &a) { return x > a.x; }
  inline bool operator >= (const modint &a) { return x >= a.x; }
  operator int() const {
    return x;
  }
  // inline void
};

typedef modint<MOD> mint;

inline mint ksm(mint a, int b, mint ret = 1) {
	for(; b; b >>= 1, a = a * a ) if(b & 1) ret = ret * a ;
	return ret;
}

constexpr int N = 2E5 + 10;

int n, m, a[N];
mint f[1 << 8][1 << 8];

// f[i][j] 表示高八位为 i 且 低八位和j没有交集的方案数

void solve() {
  cin >> n >> m;
  mint ans = 0;
  for (int i = 1; i <= n; i ++ ) {
    string s; cin >> s;
    for (int j = 0; j < m; j ++ ) if (s[j] - 'h') {
      a[i] |= 1 << j;
    }
    int x = a[i] >> 8, y = a[i] & ((1 << 8) - 1);
    mint z = 1;
    for (int i = 0; i < 1 << 8; i ++ ) if (!(x & i)) {
      z += f[i][y];
    }
    ans += z;
    for (int i = 0; i < 1 << 8; i ++ ) if (!(y & i)) {
      f[x][i] += z;
    }
  }
  cout << ans << "\n";


}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  solve();
  return 0;
}