题目意思

Solution

参考zzugzx题解

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 998244353;
const int INF = 0x3f3f3f3f;

const int N = 2e3 + 7;
vector<int> e[N];
ll n, k;
ll dp[N][N], sz[N];

void dfs(int u, int fa) {
    dp[u][1] = sz[u] = 1;
    for (auto v : e[u]) {
        if (v == fa)    continue;
        dfs(v, u);
        ll tmp = 0;
        for (int i = min(sz[v], k); i; --i)
            tmp = (tmp + dp[v][i]) % MOD;

        for (int i = min(sz[u], k); i; --i) {
            for (int j = min(sz[v], k); j; --j)
                if (i + j <= k)
                    dp[u][i + j] = (dp[u][i + j] + dp[u][i] * dp[v][j] % MOD) % MOD;

            dp[u][i] = dp[u][i] * tmp % MOD;
        }
        sz[u] += sz[v];
    }
}

int main() {
    n = read(), k = read();
    for (int i = 1; i < n; ++i) {
        int u = read(), v = read();
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1, 0);
    ll ans = 0;
    for (int i = 1; i <= k; ++i)
        ans = (ans + dp[1][i]) % MOD;
    print(ans);
    return 0;
}