https://www.luogu.com.cn/problem/P5243

大意就是給出若\(k\)棵樹，每棵樹選一條路徑，然後花\(k x\)的長度把他們連成一個環
問環的長度\(\ge Y\)的方案數

首先肯定可以先把\(Y -= kx\)
然後就變成了一個經典問題，每個樹暴力\(n_i^2\)跑出來所有的方案，路徑長度最多\(min(n_i^2,Y)\)種

然後揹包轉移，一共\(k\)個，時間複雜度為\(O(\sum n_i^2 +\sum min(Y, n_i^2)Y)\)
時間複雜度上限為\(O(nY\sqrt{Y})\)

code:

#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
#define N 2550
using namespace std;
struct edge {
    int v, nxt, c;
} e[N << 1];
int p[N], eid;
void init() {
    memset(p, -1, sizeof p);
    eid = 0;
}
void insert(int u, int v, int c) {
    e[eid].v = v;
    e[eid].c = c;
    e[eid].nxt = p[u];
    p[u] = eid ++;
}
int fa[N];
int get(int x) {
    return fa[x] == x? x : (fa[x] = get(fa[x]));
}
void merge(int x, int y) {
    x = get(x), y = get(y);
    fa[x] = y;
}
ll sum[N][N], gs[N][N];
int n, m, X, Y;
void dfs(int u, int fa, int l) {
    if(fa) (sum[get(u)][min(l, Y)] += l) %= mod, gs[get(u)][min(l, Y)] ++;
    for(int i = p[u]; i + 1; i = e[i].nxt) {
        int v = e[i].v, c = e[i].c;
        if(v == fa) continue;
        dfs(v, u, l + c);
    }
}
ll f[N][2], g[N][2];
int main() {
    init();
    scanf("%d%d%d%d", &n, &m, &X, &Y);
    for(int i = 1; i <= n; i ++) fa[i] = i;
    for(int i = 1; i <= m; i ++) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        insert(u, v, c), insert(v, u, c);
        merge(u, v);
    }
    int sz = 0;
    for(int i = 1; i <= n; i ++) if(get(i) == i) sz ++;
    Y = max(0, Y - sz * X);

    for(int i = 1; i <= n; i ++) dfs(i, 0, 0);    
    f[0][0] = 1, f[0][1] = 0;
    for(int i = 1; i <= n; i ++) if(get(i) == i) {
        for(int j = 0; j <= Y; j ++) g[j][0] = f[j][0], g[j][1] = f[j][1], f[j][0] = f[j][1] = 0;
        for(int j = 0; j <= Y; j ++) if(gs[i][j]) {
            for(int k = 0; k <= Y; k ++) if(g[k][0]) {
                int o = min(j + k, Y);
                (f[o][0] += g[k][0] * gs[i][j] % mod) %= mod;
                (f[o][1] += g[k][0] * sum[i][j] % mod + g[k][1] * gs[i][j]) %= mod;
            }
        }
    }
    // for(int i = 1; i <= n; i ++) {
    //     for(int j = 0; j <= Y; j ++) printf("%lld ", gs[i][j]); printf("\n");
    // }
    // printf("%d  %lld %lld\n", Y, f[Y][0], f[Y][1]);
    ll ans = (f[Y][1] + f[Y][0] * sz % mod * X % mod) % mod;
    for(int i = 1; i < sz; i ++) ans = ans * i % mod;
    printf("%lld", ans * ((mod + 1) / 2) % mod);
    return 0;
}