Submission #1633842


Source Code Expand

#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)n;++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int>P;

const int MAX_N = 100005;

vector<int> G[MAX_N];
double dp[MAX_N];
double dp2[MAX_N];
double ans[MAX_N];
int eda[MAX_N];

void dfs(int u,int p)
{
    int cnt = 0;
    double res = 0;
    if((int)G[u].size() == 1 && G[u][0] == p){
        dp[u] = 0;
        return;
    }
    rep(i,G[u].size()){
        if(G[u][i] != p){
            cnt++;
            dfs(G[u][i],u);
            res += dp[G[u][i]];
        }
    }
    eda[u] = cnt;
    dp[u] = res/cnt + 1;
}

void dfs2(int u,int p)
{
    rep(i,G[u].size()){
        if(G[u][i] != p){
            if(u == 0){
                if(eda[u] == 1){
                    dp2[G[u][i]] = 1;
                }else{
                    dp2[G[u][i]] = (dp[u]*eda[u]-(1+dp[G[u][i]]))/(eda[u]-1) + 1;
                }
            }else{
                dp2[G[u][i]] = dp2[u]/eda[u] + dp[u]-(1+dp[G[u][i]])/eda[u] + 1;
            }
            ans[G[u][i]] = dp2[G[u][i]]/(eda[G[u][i]]+1) + dp[G[u][i]]*eda[G[u][i]]/(eda[G[u][i]]+1);
            dfs2(G[u][i],u);
        }
    }
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    rep(i,n-1){
        int a,b;
        cin >> a >> b;
        G[a-1].pb(b-1),G[b-1].pb(a-1);
    }
    dfs(0,-1);
    ans[0] = dp[0];
    dp2[0] = 0;
    dfs2(0,-1);
    // rep(i,n){
    //     cout << i << " " << dp[i] << endl;
    // }
    rep(i,n){
        printf("%.12lf\n",ans[i]);
    }
    return 0;
}

Submission Info

Submission Time
Task D - Driving on a Tree
User kopricky
Language C++14 (GCC 5.4.1)
Score 410
Code Size 2250 Byte
Status RE
Exec Time 115 ms
Memory 16128 KB

Judge Result

Set Name Subtask1 Subtask2 Subtask3
Score / Max Score 190 / 190 220 / 220 0 / 390
Status
AC × 3
AC × 2
AC × 5
RE × 3
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 2 ms 3328 KB
sub1_in2.txt AC 3 ms 3456 KB
sub1_in3.txt AC 115 ms 16128 KB
sub2_in1.txt AC 3 ms 3456 KB
sub2_in2.txt AC 3 ms 3456 KB
sub3_in1.txt RE 98 ms 3328 KB
sub3_in2.txt RE 115 ms 6776 KB
sub3_in3.txt RE 110 ms 4864 KB