题目链接

题意：



题解：











AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
//const int mod = 1e9 + 7;
const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 3e2 + 5;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
ll n, k;
vector<int> g[2010];
ll dp[2010][2010], sz[2010];
void dfs(int u, int fa) {
    dp[u][1] = sz[u] = 1;
    for (auto v : g[u]) {
        if (v == fa) continue;
        dfs(v, u);
        ll s = 0;
        for (int i = 1; i <= min(sz[v], k); i++)
            s = (s + dp[v][i]) % mod;
        for (int i = min(k, sz[u]); i; i--) {
            for (int j =min(k, sz[v]); j; j--)
                if (i + j <= k) {
                    dp[u][i + j] = (dp[u][i + j] + dp[u][i] * dp[v][j]) % mod;
                }
            dp[u][i] = dp[u][i] * s % mod;
        }
        sz[u] += sz[v];
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    dfs(1, 0);
    ll ans = 0;
    for (int i = 1; i <= k; i++)
        ans = (ans + dp[1][i]) % mod;
    cout << ans << endl;
    return 0;
}