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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P6478 [NOI Online 2020 提高组] 游戏

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P6478 [NOI Online #2 提高组] 游戏

恰好 $k$ 个是不好求的，考虑钦定 $k$ 个后，后面的随便选取，这个可以树上背包做。那么有 $F(x) = \sum _ {k = x} ^ n \binom i k G(i)$ ，二项式反演后，得到答案 $G(x) = \sum _ {k = x} ^ n (-1) ^ {i - x} F(i)$ 。

注意，树上背包加上子树大小剪枝后就是 $O(n ^ 2)$ 。

查看代码 
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#include <cstdio>
#include <vector>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        flag |= c == '-';
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
    if (x < 0) putchar('-'), x = ~x + 1;
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e3 + 10, mod = 998244353;
vector <int> e[N];
bool col[N];
int n, sz[N][2], f[N][N], g[N];
int inv[N], fact[N], ifact[N];
void init ()
{
    inv[1] = 1;
    for (int i = 2; i < N; ++i)
        inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < N; ++i)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
    }
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
void dfs (int u, int fa)
{
    f[u][0] = 1;
    for (int v : e[u]) if (v ^ fa)
    {
        dfs(v, u);
        for (int i = 0; i <= n >> 1; ++i) g[i] = 0;
        for (int i = 0; i <= min(sz[u][0] + sz[u][1], n >> 1); ++i)
            for (int j = 0; j <= min(sz[v][0] + sz[v][1], (n >> 1) - i); ++j)
                (g[i + j] += (LL)f[u][i] * f[v][j] % mod) %= mod;
        for (int i = 0; i <= n >> 1; ++i) f[u][i] = g[i];
        sz[u][0] += sz[v][0], sz[u][1] += sz[v][1];
    }
    for (int i = n >> 1; i; --i)
        (f[u][i] += (LL)f[u][i - 1] * (sz[u][col[u] ^ 1] - i + 1) % mod) %= mod;
    ++sz[u][col[u]];
}
int main ()
{
    init();
    read(n);
    for (int i = 1; i <= n; ++i)
    {
        char c = getchar();
        while (c < '0' || c > '1') c = getchar();
        col[i] = c == '1';
    }
    for (int i = 1, a, b; i < n; ++i)
    {
        read(a, b);
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1, 0);
    for (int i = 0; i <= n >> 1; ++i)
        f[1][i] = (LL)f[1][i] * fact[(n >> 1) - i] % mod;
    for (int i = 0; i <= n >> 1; ++i)
    {
        int res = 0;
        for (int j = i; j <= n >> 1; ++j)
            (res += (j - i & 1 ? -1ll : 1ll) * C(j, i) * f[1][j] % mod) %= mod;
        write((res + mod) % mod), puts("");
    }
    return 0;
}
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