JZOJ 5251.决战

首先考虑一个 SB 状压 DP:
\(f_{i,S,j}\) 表示前 \(i\) 行,第 \(i\) 行放了哲学家的位置的集合为 \(S\),目前放了 \(j\) 个哲学家。
这个玩意复杂度是 \(O(4^c nm)\) 的,其中 \(c = 3\)

考虑设 \(F_{i,S}(x) = \sum\limits_{j=0}^{3n} f_{i,S,j} x^j\)
那么转移可以写作 \(F_{i,S}(x) = \sum\limits_{\text{It is valid to transfer from } T \text{ to } S \text.} x^{|S|} F_{i-1,T}(x)\)

考虑基于 \(F\) 的点值表达式进行 DP,设 \(F_{i,k,S}\)\(F_{i,S}(\omega_{3n}^k)\),那么便可把上述式子写成点值相乘并相加的形式。
然而复杂度还是 \(O(4^c nm)\) 的,并无卵用(

真的无卵用吗?注意到这样子我们就让 \(S\) 这一维的转移相对独立了,并且 \(S\) 十分的小,考虑用矩阵乘法优化这一维的转移。
复杂度为 \(O(8^c n \log n)\)
(注意多项式的次数界应为 \(3n\),因为我们无法在转移的同时把点值 \(\bmod x^m\)。)

代码:

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#include <cstdio>
#include <cstring>
#include <algorithm>
#define add(a,b) (a + b >= mod ? a + b - mod : a + b)
#define dec(a,b) (a < b ? a - b + mod : a - b)
using namespace std;
const int N = 2500;
const int M = 7500;
const int K = 3;
const int S = 8;
const int mod = 998244353;
const int G = 3;
int n,m,full = (1 << K) - 1,a[K + 5];
int up[(1 << K) + 5],down[(1 << K) + 5],valid[(1 << K) + 5];
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
namespace NTT
{
const int N = 15000;
struct poly
{
int a[N + 5];
inline const int &operator[](int x) const
{
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear(int x = 0)
{
memset(a + x,0,(N - x + 1) << 2);
}
} f;
int n,lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init(int len)
{
for(n = 1;n < len;n <<= 1);
for(register int i = 2;i <= n;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / n);
rt[n >> 1] = 1;
for(register int i = (n >> 1) + 1;i <= n;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (n >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= n;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[n] = fpow(fac[n],mod - 2);
for(register int i = n;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= n;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
inline void ntt(poly &a,int type,int n)
{
type == -1 && (reverse(a.a + 1,a.a + n),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
}
using NTT::init;
using NTT::ntt;
struct Matrix
{
int a[S + 5][S + 5];
inline void clear()
{
memset(a,0,sizeof a);
}
inline Matrix()
{
memset(a,0,sizeof a);
}
inline Matrix(int)
{
memset(a,0,sizeof a);
for(register int i = 0;i <= full;++i)
a[i][i] = 1;
}
inline const int *operator[](const int &x) const
{
return a[x];
}
inline int *operator[](const int &x)
{
return a[x];
}
inline Matrix operator*(const Matrix &o) const
{
Matrix ret;
for(register int i = 0;i <= full;++i)
for(register int j = 0;j <= full;++j)
for(register int k = 0;k <= full;++k)
ret[i][j] = (ret[i][j] + (long long)a[i][k] * o[k][j]) % mod;
return ret;
}
} mat;
inline Matrix fpow(Matrix a,int b)
{
Matrix ret(1);
for(;b;b >>= 1)
(b & 1) && (ret = ret * a,1),a = a * a;
return ret;
}
int pw[4];
int dp(int w)
{
int ret = 0;
pw[0] = 1;
for(register int i = 1;i <= K;++i)
pw[i] = (long long)pw[i - 1] * w % mod;
mat.clear();
for(register int S = 0;S <= full;++S)
for(register int T = 0;T <= full;++T)
if(valid[S] && valid[T] && !(up[T] & S) && !(down[S] & T))
mat[S][T] = pw[__builtin_popcount(T)];
mat = fpow(mat,n);
for(register int i = 0;i <= full;++i)
ret = add(ret,mat[0][i]);
return ret;
}
int main()
{
scanf("%d%d",&n,&m);
if(m > 3 * n)
{
puts("0");
return 0;
}
int x;
for(register int i = 1;i <= K;++i)
for(register int j = 1;j <= K;++j)
scanf("%d",&x),a[i] |= x << j - 1;
for(register int i = 0;i <= full;++i)
{
valid[i] = 1;
for(register int j = 1;j <= K;++j)
if(i & (1 << j - 1))
up[i] |= a[1] << 1 >> K - j,down[i] |= a[3] << 1 >> K - j,
valid[i] &= !((i ^ (1 << j - 1)) & (a[2] << 1 >> K - j));
up[i] &= full,down[i] &= full;
}
NTT::init(n * 3);
for(register int i = 0;i < NTT::n;++i)
NTT::f[i] = dp(fpow(G,(mod - 1) / NTT::n * i));
ntt(NTT::f,-1,NTT::n);
printf("%d\n",NTT::f[m]);
}