use epsilon 4.6e-9

7 4
6 4
5 3
7 5
8 6

1 1
1 1

you print Ask Shiftman for help. once again, although the correct answer is Perfect

1 0 0 0

just imagine a tree

super incomprehensible, in general the tests are about some kind of accuracy, you need to make 100 parcels and try different epsilons everywhere to pass these tests, it pissed me off

hypercircle does not interfere with the thread

2
abbb
abbb

dont forgot about solo points

Try this one
0 0
1 1
1 0
2 -1

Ans: 0

Try:
101 203 1000

Or:
1 3 1000

3 2
3
3 3 3

I think this test should break some AC solutions:

443 -39 457 33
386 -773 561 127

Correct answer is -1, but I think some AC solutions can print 0

read the tags before you decide, and think a few times before you start

wa5 - check how you calculate time when change direction at the station

Test:
12
1 2 45 64 31 23 128 61 2 2 7
8
7 3 5

ans: 805


Test:
4
7 6 3
3
12 3 4
ans : 60

Good luck!

stupid error

Here's my wrong solution, can sb view the code and help me?
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> pos;
vector<vector<int>> in;
vector<vector<int>> out;
vector<int> ans;

void lets_go(int i)
{
    for (int j = 0; j < in[i].size(); j++)
    {
        if (pos[in[i][j]] == 0)
        {
            lets_go(in[i][j]);
        }
    }
    pos[i] = 1;
    ans.push_back(i);
    for (int j = 0; j < out[i].size(); j++)
    {
        if (pos[out[i][j]] == 0)
        {
            lets_go(out[i][j]);
        }
    }
}

int main()
{
    int n;
    cin >> n;
    in.resize(n);
    out.resize(n);
    pos.resize(n, 0);
    for (int i = 0; i < n; i++)
    {
        int a;
        cin >> a;
        while (a != 0)
        {
            if (a != 0)
            {
                in[a - 1].push_back(i);
                out[i].push_back(a - 1);
                cin >> a;
            }
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (pos[i] == 0)
        {
            lets_go(i);
        }
    }
    for (int i = 0; i < n; i++)
    {
        cout << ans[i] + 1 << " ";
    }
}

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