Submission #2095709

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Source Code Expand

#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define int ll
using VI = vector<int>;
using VVI = vector<VI>;
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()
#define PB push_back

const ll LLINF = (1LL<<60);
const int INF = (1LL<<30);
const int MOD = 1000000007;

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<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a){
  out<<'('<<a.first<<','<<a.second<<')';
  return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
  out<<'[';
  REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';}
  out<<']';
  return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

VI g[100010];
signed main(void)
{
  int n;
  cin >> n;
  REP(i, n-1) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].PB(b);
    g[b].PB(a);
  }

  vector<double> prob(n);
  function<double(int,int)> dfs1 = [&](int v, int p) -> double {
    if(p != -1 && g[v].size() == 1) return 0;
    int cnt = 0;
    double ret = 0;
    for(int &i: g[v]) {
      if(i == p) continue;
      cnt++;
      ret += dfs1(i, v) + 1;
    }
    ret /= cnt;
    return prob[v] = ret;
  };

  dfs1(0, -1);
  // cout << prob << endl;

  vector<double> ans(n);
  function<void(int,double,int)> dfs2 = [&](int v, double d_par, int p) {
    // cout << v << " " << d_par << " " << p << endl;
    // vector<PII> d_child;
    // d_child.emplace_back(0, -1);
    double ret = 0;
    for(int &e : g[v]) {
      if(e == p) {
        ret += d_par + 1;
      } else {
        ret += prob[e] + 1;
      }
    }
    // sort(d_child.rbegin(), d_child.rend());
    ans[v] = ret / g[v].size();
    for(int &e : g[v]) {
      if(e == p) continue;
      double nxt = g[v].size() == 1 ? 0 : (ret - prob[e] - 1) / (g[v].size() - 1);
      dfs2(e, nxt, v);
    }
  };

  dfs2(0, 0, -1);
  REP(i, n) {
    cout << fixed << setprecision(9) << ans[i] << endl;
  }

  return 0;
}

Submission Info

Judge Result