19-9-25-3

LOJ 2340

Solution

考虑状压dpdp

FSF_{S}表示SS集合的答案,WSW_S表示SS集合中的点权和,GS=[S is valid]WSG_S = [S~is~valid]W_S,枚举最后加进去的集合TT,那么有

FS=TS,TGTWSFST=1WSTS,TGTFST \begin{aligned} F_S &= \sum_{T\subseteq S, T\ne \emptyset} \frac{G_{T}}{W_S}\cdot F_{S - T}\\ &= \frac{1}{W_S}\sum_{T\subseteq S, T\ne \emptyset} G_{T}\cdot F_{S - T}\\ \end{aligned}

显然可以子集卷积

具体在判断一个集合是否合法的时候注意要判所有度数为偶数和联通性

然后这里的子集卷积必须每次都IDWT出来,除掉WSW_S之后再DWT回去。。。

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
T sum = 0, fl = 1; char ch = getchar();
for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
return sum * fl;
}

inline void proc_status ()
{
ifstream t ("/proc/self/status");
cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 21 + 5, Maxs = (1 << 21) + 5;
const int Mod = 998244353;

namespace MATH
{
inline void Add (int &a, int b) { if ((a += b) >= Mod) a -= Mod; }

inline int Pow (int a, int b)
{
int ans = 1;
for (int i = b; i; i >>= 1, a = (LL) a * a % Mod) if (i & 1) ans = (LL) ans * a % Mod;
return ans;
}
}

using namespace MATH;

int N, M, P, ALL;
int G[Maxn], W[Maxs], IW[Maxs];
int f[Maxn][Maxs], g[Maxn][Maxs];

namespace DSU
{
int fa[Maxn];
inline void init () { for (int i = 0; i < N; ++i) fa[i] = i; }
inline int get_fa (int x) { return fa[x] == x ? x : fa[x] = get_fa (fa[x]); }
inline void link (int x, int y) { fa[get_fa (x)] = get_fa (y); }
}

inline int check (int S)
{
static int deg[Maxn];
for (int i = 0; i < N; ++i) deg[i] = 0;
DSU :: init ();

for (int i = 0; i < N; ++i) if (S & (1 << i))
for (int j = i + 1; j < N; ++j) if ((S & (1 << j)) && (G[i] & (1 << j)))
DSU :: link (i, j), ++deg[i], ++deg[j];

int cnt = 0;
for (int i = 0; i < N; ++i) if ((1 << i) & S)
{
if (DSU :: fa[i] == i) ++cnt;
if (deg[i] & 1) return 1;
}
if (cnt > 1) return 1;
return 0;
}

inline void Init ()
{
ALL = (1 << N) - 1;
for (int i = 1; i <= ALL; ++i)
{
int p = i & (-i);
W[i] = (W[i ^ p] + W[p]) % Mod;
}

for (int i = 0; i <= ALL; ++i)
{
W[i] = Pow (W[i], P);
IW[i] = Pow (W[i], Mod - 2);

int len = __builtin_popcount (i);
if (check (i)) g[len][i] = W[i];
}
}

inline void DWT (int *A, int n, int fg)
{
for (int mid = 1; mid < n; mid <<= 1)
for (int i = 0, len = mid << 1; i < n; i += len)
for (int j = i; j < i + mid; ++j)
{
int x = A[j], y = A[j + mid];
if (!fg) A[j + mid] = (x + y) % Mod;
else A[j + mid] = (y - x + Mod) % Mod;
}
}

inline void Solve ()
{
Init ();

f[0][0] = 1;
DWT (f[0], 1 << N, 0);
for (int i = 0; i <= N; ++i) DWT (g[i], 1 << N, 0);

for (int i = 1; i <= N; ++i)
{
for (int j = 0; j < i; ++j)
for (int S = 0; S <= ALL; ++S)
Add (f[i][S], (LL) f[j][S] * g[i - j][S] % Mod);

DWT (f[i], 1 << N, 1);

for (int S = 0; S <= ALL; ++S) f[i][S] = (LL) f[i][S] * IW[S] % Mod;

DWT (f[i], 1 << N, 0);
}

DWT (f[N], 1 << N, 1);
printf("%d\n", f[N][ALL]);
}

inline void Input ()
{
N = read<int>(), M = read<int>(), P = read<int>();
for (int i = 1; i <= M; ++i)
{
int x = read<int>() - 1, y = read<int>() - 1;
G[x] |= (1 << y);
G[y] |= (1 << x);
}
for (int i = 0; i < N; ++i) W[1 << i] = read<int>();
}

int main()
{

#ifdef hk_cnyali
freopen("A.in", "r", stdin);
freopen("A.out", "w", stdout);
#endif

Input ();
Solve ();

return 0;
}