「九省联考 2018」秘密袭击 （树形DP）（生成函数）（数数）（拉格朗日插值）（线段树合并）

​ 传送门​​

• 题解：
  考虑每个第大贡献多少个集合
  
  那么我们可以枚举每一个，然后算出第大的个数
  这个个数等价于的至少有个的个数
  那么我们可以令表示到有个的个数
  写成生成函数，求的是
  本来以为要然后分治的，好像可以两个，突然发现模数不对
  其实可以考虑求出个点值，然后拉个朗日插值，其中拉个朗日插值的复杂度可以做到
  暴力求个点值的复杂度是，但是我们可以把枚举这一步省掉
  用线段树维护每一个的答案，那么初始化就是一段赋 1 一段赋
  维护一个系数类表示把变成就可以表示所有转移
  并支持结合律
  考虑线段树合并把每个值都搞上去，当合并到一棵树的叶子结点也就是下面一个区间全部是 0 的时候，整个区间对当前的贡献可以通过这个系数类对区间打一个乘法标记，这样就支持合并了
  复杂度

#include<bits/stdc++.h>
#define cs const
using namespace std;
int read(){
  int cnt = 0, f = 1; char ch = 0;
  while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1; }
  while(isdigit(ch)) cnt = cnt*10 + (ch-'0'), ch = getchar();
  return cnt * f;
}
cs int N = 1680;
cs int Mod = 64123;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int ksm(int a, int b){ int ans = 1; for(;b;b>>=1,a=mul(a,a)) if(b&1) ans = mul(ans, a); return ans; }
void Add(int &a, int b){ a = add(a, b); }
void Mul(int &a, int b){ a = mul(a, b); }
void Dec(int &a, int b){ a = dec(a, b); }
int sgn(int a){ return a & 1 ? Mod - 1 : 1; }
int n, k, W, d[N];
vector<int> G[N];
#define pb push_back
#define poly vector<int>
int rt[N]; 
struct Coef{
  int a, b, c, d; 
  Coef(int _a=0, int _b=0, int _c=0, int _d=0){ a=_a; b=_b; c=_c; d=_d; }
  Coef operator * (cs Coef &A){
  return Coef(mul(a,A.a),add(mul(b,A.a),A.b),add(mul(a,A.c),c),add(mul(b,A.c),add(d,A.d)));
  }
  void operator *= (cs Coef &A){ *this = *this * A; }
};
namespace SGT{
  cs int N = ::N * 200;
  int nd; int ls[N], rs[N]; Coef vl[N];
  int newnode(){ int x = ++nd; ls[x] = rs[x] = 0; vl[x] = Coef(1,0,0,0); return x; }
  #define mid ((l+r)>>1)
  void down(int x){
    if(!ls[x]) ls[x] = newnode();
    if(!rs[x]) rs[x] = newnode();
    vl[ls[x]] *= vl[x];
    vl[rs[x]] *= vl[x];
    vl[x] = Coef(1,0,0,0);
  }
  void modify(int &x, int l, int r, int L, int R, Coef coe){
    if(!x) x = newnode(); if(L<=l && r<=R){ vl[x] *= coe; return; } down(x); 
    if(L<=mid) modify(ls[x], l, mid, L, R, coe);
    if(R>mid) modify(rs[x], mid+1, r, L, R, coe);
  }
  void merge(int &x, int y){
    if(!x || !y){ x |= y; return; }
    if(!ls[x] && !rs[x]) swap(x, y);
    if(!ls[y] && !rs[y]){
      vl[x] *= Coef(vl[y].b,0,0,0);
      vl[x] *= Coef(1,0,0,vl[y].d);
      return;
    } down(x); down(y);
    merge(ls[x], ls[y]);
    merge(rs[x], rs[y]);
  }
  int query(int x, int l, int r){
    if(l == r) return vl[x].d; down(x);
    return add(query(ls[x],l,mid), query(rs[x],mid+1,r));
  }
}
void dfs(int u, int fa, int c){
  SGT::modify(rt[u],1,W,1,W,Coef(0,1,0,0));
  for(int v : G[u]) if(v ^ fa){
    dfs(v, u, c);
    SGT::merge(rt[u], rt[v]);
  } 
  SGT::modify(rt[u],1,W,1,d[u],Coef(c,0,0,0));
  SGT::modify(rt[u],1,W,1,W,Coef(1,0,1,0));
  SGT::modify(rt[u],1,W,1,W,Coef(1,1,0,0));
}
int y[N], inv[N];
poly operator * (poly a, poly b){
  int deg = a.size() + b.size() - 1; poly c(deg,0);
  for(int i = 0; i < a.size(); i++)
  for(int j = 0; j < b.size(); j++)
  Add(c[i+j], mul(a[i],b[j])); return c;
}
poly operator * (poly a, int coe){
  for(int i = 0; i < a.size(); i++) Mul(a[i],coe); 
  return a;
}
poly operator + (poly a, poly b){
  int deg = max(a.size(),b.size()); a.resize(deg); b.resize(deg);
  for(int i = 0; i < deg; i++) Add(a[i],b[i]); return a;
}
poly operator / (poly a, poly b){
  int deg = a.size() - b.size() + 1; poly c(deg, 0);
  for(int i = (int)a.size() - 1; i > 0; i--){
    c[i-1] = a[i]; Dec(a[i-1], mul(b[0], a[i]));
  } return c;
}
int work(int n){
  inv[0] = inv[1] = 1;
  for(int i = 2; i <= n; i++) inv[i] = mul(Mod-Mod/i, inv[Mod%i]);
  for(int i = 2; i <= n; i++) Mul(inv[i],inv[i-1]);
  poly f, g; f.pb(1); g.pb(0); g.pb(1);
  for(int i = 1; i <= n; i++) Dec(g[0],1), f = f * g;
  poly ans;
  for(int i = 1; i <= n; i++){
    int coe = mul(mul(sgn(n-i),y[i]),mul(inv[i-1],inv[n-i]));
    poly res; res.pb(dec(0,i)); res.pb(1); 
    ans = ans + (f / res) * coe;
  }
  int sm = 0;
  for(int i = k; i < ans.size(); i++) Add(sm, ans[i]);
  return sm; 
}
int main(){
  n = read(), k = read(), W = read();
  for(int i = 1; i <= n; i++) d[i] = read();
  for(int i = 1, x, y; i < n; i++) 
    x = read(), y = read(), G[x].pb(y), G[y].pb(x);
  for(int c = 1; c <= n + 1; c++){
    dfs(1,0,c);
    y[c] = SGT::query(rt[1],1,W); SGT::nd = 0;
    memset(rt, 0, sizeof(int)*(n+1));
  } cout << work(n+1); return 0;
}