結果

問題 No.196 典型DP (1)
コンテスト
ユーザー fumiphys
提出日時 2019-09-29 23:33:01
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 23 ms / 2,000 ms
コード長 3,765 bytes
コンパイル時間 1,618 ms
コンパイル使用メモリ 172,864 KB
実行使用メモリ 26,752 KB
最終ジャッジ日時 2024-10-03 04:43:45
合計ジャッジ時間 3,198 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

vector<int> edge[2010], child[2010];
int s[2010];
void dfs(int i){
  s[i] = 1;
  for(auto e: edge[i]){
    if(s[e] == 0){
      dfs(e);
      s[i] += s[e];
      child[i].pb(e);
    }
  }
}

ll dp[2010][2010];
void calc(int i){
  for(auto e: child[i])calc(e);
  vector<ll> pre(s[i] + 1, 0), nex(s[i] + 1, 0);
  pre[0] = 1;
  int acc = 0;
  for(auto e: child[i]){
    for(int j = 0; j <= acc + s[e]; j++)nex[j] = 0;
    for(int j = 0; j <= acc; j++){
      for(int k = 0; k <= s[e]; k++){
        (nex[j+k] += pre[j] * dp[e][k] % mod) %= mod;
      }
    }
    acc += s[e];
    swap(pre, nex);
  }
  pre[s[i]] = (pre[s[i]] + 1) % mod;
  rep(k, sz(pre))dp[i][k] = pre[k];
}

int main(int argc, char const* argv[])
{
  int n, k; cin >> n >> k;
  rep(i, n - 1){
    int a, b; cin >> a >> b;
    edge[a].pb(b);
    edge[b].pb(a);
  }
  dfs(0);
  calc(0);
  cout << dp[0][k] << endl;
  return 0;
}
0