Codeforces Round #518 (Div. 1) B. Multihedgehog

問題ページ

考えたこと

• 各頂点を根として判定をするO(N^2)はできそう
• 全方位木DPかなあとか考える
• 冷静になると根になりそうなのはグラフの中心しかない
• 木の直径はdouble-sweepで求められるので直径を構成するパスの端点から距離が等しい頂点が中心
• 木の中心からdfsしてそのグラフがk-multihedgehogかどうか判定する

#include <bits/stdc++.h>
 
using namespace std;
using ll = long long;
// #define int ll
using PII = pair<int, int>;
 
#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
 
template<typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }
 
template<typename T> vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts) { 
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type
fill_v(T &t, const V &v) { t=v; }
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t, const V &v ) { for(auto &e:t) fill_v(e,v); }
 
template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')'; return out;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
  for(T& x: vec) {is >> x;} return is;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'['; for(T i: a) {out<<i<<',';} out<<']'; return out;
}
 
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; // DRUL
const int INF = 1<<30;
const ll LLINF = 1LL<<60;
ll MOD = 1000000007;

signed main(void)
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  ll n, k;
  cin >> n >> k;
  vector<vector<ll>> g(n);
  REP(i, n-1) {
    ll u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }

  ll mode;
  auto d = make_v<ll>(3, n);
  function<void(ll,ll,ll)> dfs = [&](ll v, ll p, ll dist) {
    d[mode][v] = dist;
    for(auto to: g[v]) {
      if(to == p) continue;
      dfs(to, v, dist + 1);
    }
  };

  mode = 0;
  dfs(0, -1, 0);
  ll ma = -INF, idx1 = -1;
  REP(i, n) {
    if(ma < d[0][i]) {
      ma = d[0][i];
      idx1 = i;
    }
  }

  mode = 1;
  dfs(idx1, -1, 0);
  ma = -INF; ll idx2 = -1;
  REP(i, n) {
    if(ma < d[1][i]) {
      ma = d[1][i];
      idx2 = i;
    }
  }

  if(ma%2) {
    cout << "No" << endl;
    return 0;
  }

  mode = 2;
  dfs(idx2, -1, 0);
  ll root = -1;
  REP(i, n) {
    if(d[1][i] == ma/2 && d[2][i] == ma/2) {
      root = i;
    }
  }

  function<ll(ll,ll)> dfs2 = [&](ll v, ll p) {
    ll ret = -2;
    for(auto to: g[v]) {
      if(to == p) continue;
      ll tmp = dfs2(to, v);
      if(ret == -2) ret = tmp;
      else if(ret != tmp) ret = -1;
    }
    if(g[v].size()==1) ret = 0;
    else if(p!=-1 && g[v].size()<=3) ret = -1;
    else if(p==-1 && g[v].size()<=2) ret = -1;
    else ret++;
    return ret;
  };

  ll ret = dfs2(root, -1);
  if(ret == k) cout << "Yes" << endl;
  else cout << "No" << endl;

  return 0;
}