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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#593227#6634. Central SubsetKandarp08RE 0ms3596kbC++203.1kb2024-09-27 12:47:172024-09-27 12:47:17

Judging History

你现在查看的是最新测评结果

  • [2024-09-27 12:47:17]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:3596kb
  • [2024-09-27 12:47:17]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define MOD 1000000007
#define HASH_MOD1 998244353
#define HASH_MOD2 1000000009
#define HASH 31
#define ll long long
#define int long long

typedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;

template <typename T>
istream& operator>> (istream& in, vector<T>& a)
{
    for (int i = 0; i < a.size(); ++i)
        in >> a[i];

    return in;
}

template <typename T>
ostream& operator<< (ostream& out, vector<T>& a)
{
    for (int i = 0; i < a.size(); ++i)
        out << a[i] << " ";

    out << "\n";

    return out;
}

template <typename T>
ostream& operator<< (ostream& out, set<T>& s)
{
    for (int x : s)
        out << x << " ";

    out << "\n";

    return out;
}

template <typename T>
ostream& operator<< (ostream& out, vector<pair<T, T>>& v)
{
    for (int i = 0; i < v.size(); ++i)
        out << v[i].first << " " << v[i].second << "\n";

    return out;
}

ll modinv(ll a)
{
    return a <= 1 ? a : MOD - (MOD / a) * modinv(MOD % a) % MOD; 
}

ll binpow(ll x, ll y)
{
    ll ans = 1;

    while (y > 0)
    {
        if (y % 2 == 1)
            ans *= x;

        x = x * x;
        y = y / 2;

        x %= MOD;
        ans %= MOD;
    }

    return ans;
}

void init(vector<int> adj[], vector<int> children[], bool visited[], int u)
{
    visited[u] = true;

    for (int v : adj[u])
    {
        if (!visited[v])
        {
            children[u].push_back(v);
            init(adj, children, visited, v);
        }
    }
}

int dfs(vector<int> children[], vector<int>& taken, int u, int d, int sq)
{
    int mx_depth = d;

    //cout << "u = " << u << " depth = " << d << "\n";

    for (int v : children[u])
    {
        int res = dfs(children, taken, v, d + 1, sq);

        //cout << "u = " << u << " v = " << v << " res = " << res << "\n";

        if (res == -1)
            continue;

        if (res - d == sq)
        {
            taken.push_back(u);
            return -1;
        }

        mx_depth = max(mx_depth, res);
    }

    return mx_depth;
}

void solve()
{
    int n, m, i;
    cin >> n >> m;

    vector<int> adj[n], children[n];
    bool visited[n] = {false};

    for (i = 0; i < m; ++i)
    {
        int u, v;
        cin >> u >> v;

        adj[u - 1].push_back(v - 1);
        adj[v - 1].push_back(u - 1);
    }    

    init(adj, children, visited, 0);
    vector<int> taken;

    int sq = (int)ceil(sqrt(n));

    dfs(children, taken, 0, 0, sq);
    
    if (taken.size() < sq && taken.back() != 0)
        taken.push_back(0);

    cout << taken.size() << "\n";

    for (i = 0; i < taken.size(); ++i)
        cout << taken[i] + 1 << " ";

    cout << "\n";
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;

    while (t--)
    {
        solve();
    }
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3596kb

input:

2
4 3
1 2
2 3
3 4
6 7
1 2
2 3
3 1
1 4
4 5
5 6
6 4

output:

2
2 1 
1
1 

result:

ok correct (2 test cases)

Test #2:

score: -100
Runtime Error

input:

10000
15 14
13 12
5 4
9 8
11 12
15 14
10 9
14 13
2 3
2 1
6 5
10 11
3 4
7 6
8 7
6 5
2 1
2 4
4 6
2 3
3 5
10 9
8 3
9 4
5 6
5 10
3 2
5 4
2 7
1 2
4 3
2 1
2 1
2 1
2 1
9 8
9 8
5 4
1 2
6 5
3 4
3 2
7 8
7 6
2 1
1 2
14 13
3 10
5 6
2 9
11 4
2 3
2 1
8 7
13 6
5 4
5 12
6 7
4 3
7 14
16 15
2 3
2 1
6 10
6 9
6 4
9 11
...

output:


result: