結果

問題 No.196 典型DP (1)
ユーザー tanimani364
提出日時 2022-04-21 10:26:49
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 47 ms / 2,000 ms
コード長 4,447 bytes
コンパイル時間 1,799 ms
コンパイル使用メモリ 202,192 KB
最終ジャッジ日時 2025-01-28 19:22:23
ジャッジサーバーID
（参考情報） 		judge3 / judge5
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 41
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//#include <atcoder/all>
using namespace std;
#define rep(i, a) for (int i = (int)0; i < (int)a; ++i)
#define repl(i, a) for (long long i = (long long)0; i < (long long)a; ++i)
#define rrep(i, a) for (int i = (int)a; i > -1; --i)
#define REP(i, a, b) for (int i = (int)a; i < (int)b; ++i)
#define REPL(i, a, b) for (long long i = (long long)a; i < (long long)b; ++i)
#define RREP(i, a, b) for (int i = (int)a; i > b; --i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll
#define fi first
#define se second
using ll = long long;
constexpr ll mod = 1e9 + 7;
constexpr ll mod_998244353 = 998244353;
constexpr ll INF = 1LL << 60;

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

// using lll = boost::multiprecision::cpp_int;
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}

ll mypow(ll x, ll n, const ll &p = -1) {  // x^nをmodで割った余り

  if (p != -1) {
    x = (x % p + p) % p;
  }
  ll ret = 1;
  while (n > 0) {
    if (n & 1) {
      if (p != -1)
        ret = (ret * x) % p;
      else
        ret *= x;
    }
    if (p != -1)
      x = (x * x) % p;
    else
      x *= x;
    n >>= 1;
  }
  return ret;
}

template <typename T>
struct myrand {
  random_device seed;
  mt19937 mt;
  myrand() : mt(seed()) {}
  T operator()(T a, T b) {  //[a,b)
    uniform_int_distribution<T> dist(a, b - 1);
    return dist(mt);
  }
};

//using namespace atcoder;

//------------------------
//-----------------------
//------------------------
//------------------------
//------------------------

template<int mod>
struct Modint{
    int x;
    Modint():x(0){}
    Modint(int64_t y):x((y%mod+mod)%mod){}

    Modint &operator+=(const Modint &p){
			if((x+=p.x)>=mod)
				x -= mod;
			return *this;
		}

		Modint &operator-=(const Modint &p){
			if((x+=mod-p.x)>=mod)
				x -= mod;
			return *this;
		}

		Modint &operator*=(const Modint &p){
			x = (1LL * x * p.x) % mod;
			return *this;
		}

		Modint &operator/=(const Modint &p){
			*this *= p.inverse();
			return *this;
		}

		Modint operator-() const { return Modint(-x); }
		Modint operator+(const Modint &p) const{
			return Modint(*this) += p;
		}
		Modint operator-(const Modint &p) const{
			return Modint(*this) -= p;
		}
		Modint operator*(const Modint &p) const{
			return Modint(*this) *= p;
		}
		Modint operator/(const Modint &p) const{
			return Modint(*this) /= p;
		}

		bool operator==(const Modint &p) const { return x == p.x; }
		bool operator!=(const Modint &p) const{return x != p.x;}

		Modint inverse() const{//非再帰拡張ユークリッド
			int a = x, b = mod, u = 1, v = 0;
			while(b>0){
				int t = a / b;
				swap(a -= t * b, b);
				swap(u -= t * v, v);
			}
			return Modint(u);
		}

		Modint pow(int64_t n) const{//繰り返し二乗法
			Modint ret(1), mul(x);
			while(n>0){
				if(n&1)
					ret *= mul;
				mul *= mul;
				n >>= 1;
			}
			return ret;
		}

		friend ostream &operator<<(ostream &os,const Modint &p){
			return os << p.x;
		}
};

using modint = Modint<mod>;
using modint2= Modint<mod_998244353>;

vector<modint> dp[2000][3];
void solve() {
  int n,k;
  cin>>n>>k;
  vector<vector<int>>g(n);
  rep(i,n-1){
    int a,b;
    cin>>a>>b;
    g[a].eb(b);
    g[b].eb(a);
  }

  auto rec=[&](auto self,int v,int p=-1)->void{//0:白,1:黒
    rep(i,2)dp[v][i]=vector<modint>(2);
    dp[v][0][0]=1;
    dp[v][1][1]=1;
    for(int &x:g[v]){
      if(x==p)continue;
      self(self,x,v);
      int sv=dp[v][0].size();
      int sx=dp[x][0].size();
      vector<modint>nxdp[2];
      rep(i,2)nxdp[i].resize(sv+sx-1);

      rep(i,sv){
        rep(j,sx){
          if(i+j>=sv+sx-1)continue;
          nxdp[0][i+j]+=dp[v][0][i]*dp[x][0][j];
          nxdp[0][i+j]+=dp[v][0][i]*dp[x][1][j];
          nxdp[1][i+j]+=dp[v][1][i]*dp[x][1][j];
        }
      }
      rep(i,2)swap(nxdp[i],dp[v][i]);
    }
  };
  rec(rec,0);
  modint ans=dp[0][0][k]+dp[0][1][k];
  cout<<ans<<"\n";
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(15);
  solve();
  return 0;
}
0