Submission #2095709
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
Submission Time |
|
Task |
D - Driving on a Tree |
User |
ferin_tech |
Language |
C++14 (GCC 5.4.1) |
Score |
410 |
Code Size |
2397 Byte |
Status |
RE |
Exec Time |
376 ms |
Memory |
16896 KB |
Judge Result
Set Name |
Subtask1 |
Subtask2 |
Subtask3 |
Score / Max Score |
190 / 190 |
220 / 220 |
0 / 390 |
Status |
|
|
|
Set Name |
Test Cases |
Subtask1 |
sub1_in1.txt, sub1_in2.txt, sub1_in3.txt |
Subtask2 |
sub2_in1.txt, sub2_in2.txt |
Subtask3 |
sub1_in1.txt, sub1_in2.txt, sub1_in3.txt, sub2_in1.txt, sub2_in2.txt, sub3_in1.txt, sub3_in2.txt, sub3_in3.txt |
Case Name |
Status |
Exec Time |
Memory |
sub1_in1.txt |
AC |
3 ms |
2560 KB |
sub1_in2.txt |
AC |
6 ms |
2688 KB |
sub1_in3.txt |
AC |
376 ms |
16896 KB |
sub2_in1.txt |
AC |
5 ms |
2688 KB |
sub2_in2.txt |
AC |
5 ms |
2688 KB |
sub3_in1.txt |
RE |
98 ms |
2560 KB |
sub3_in2.txt |
RE |
145 ms |
6516 KB |
sub3_in3.txt |
RE |
130 ms |
4096 KB |