Submission #249255

#	Time	Username	Problem	Language	Result	Execution time	Memory
249255		Vimmer	Zagonetka (COI18_zagonetka)	C++14	100 / 100	93 ms	2556 KiB

zagonetka

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("fast-math")
//#pragma GCC optimize("no-stack-protector")

#define F first
#define S second
#define sz(x) int(x.size())
#define pb push_back
#define pf push_front
#define N 100010
#define M ll(1e9 + 7)
#define inf 1e9 + 1e9

using namespace std;
//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef short int si;
typedef array <int, 6> a6;

//typedef tree <int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

map <vector <int>, bool> mp;

bool checker(vector <int> &g)
{
    if (mp.find(g) != mp.end()) return mp[g];

    cout << "query ";

    for (auto it : g) cout << it << " ";

    cout << endl;

    bool ans;

    cin >> ans;

    mp[g] = ans;

    return ans;
}


vector <int> ans;

int p[150], pos[150], cur = 1, n;

bool gd[150][150], bn[150][150], mk[150];

void dfs(int v)
{
    if (mk[v]) return;

    mk[v] = 1;

    for (int i = 0; i < n; i++)
    {
        if (i == v) continue;

        if (bn[i][v]) dfs(i);
    }

    ans[v] = cur++;
}

void rec(int v)
{
    if (mk[v]) return;

    mk[v] = 1;

    for (int i = 0; i < n; i++)
    {
        if (i == v) continue;

        if (bn[v][i]) rec(i);
    }

    ans[v] = cur--;
}
int main()
{
    //freopen("input.txt", "r", stdin); freopen("output4.txt", "w", stdout);

    ios_base::sync_with_stdio(0); istream::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    cin >> n;

    for (int i = 0; i < n; i++)
    {
        cin >> p[i];

        pos[p[i]] = i;

        gd[p[i]][p[i]] = 1;
    }

    for (int len = 1; len <= n; len++)
      for (int i = 1; i <= n; i++)
      {
            int nxt = i + len;

            if (nxt > n) continue;

            bool fnd = 0;

            for (int md = i + 1; md < nxt; md++)
              if (gd[i][md] && gd[md][nxt]) {fnd = 1; break;}

            if (fnd)
            {
                gd[i][nxt] = 1;

                continue;
            }

            vector <int> v, gr; v.clear(); gr.resize(n);

            for (int j = 0; j < n; j++) gr[j] = p[j];

            for (int md = i; md <= nxt; md++)
                if (gd[i][md] || gd[md][nxt]) v.pb(md);

            reverse(v.begin(), v.end());

            int j = 0;

            for (int md = nxt; md >= i; md--)
                if (gd[i][md]) gr[pos[md]] = v[j++];

            for (int md = nxt; md >= i; md--)
                if (gd[md][nxt]) gr[pos[md]] = v[j++];

            gd[i][nxt] = !checker(gr);
      }

    for (int i = 1; i <= n; i++)
        for (int j = i + 1; j <= n; j++)
          bn[pos[i]][pos[j]] = gd[i][j];

    ans.resize(n);

    for (int i = 0; i < n; i++)
        dfs(i);

    cout << "end" << endl;

    for (auto it : ans) cout << it << " "; cout << endl;

    cur = n;

    memset(mk, 0, sizeof(mk));

    ans.resize(n);

    for (int i = 0; i < n; i++)
        rec(i);

    for (auto it : ans) cout << it << " "; cout << endl;
}

Compilation message (stderr)

zagonetka.cpp: In function 'int main()':
zagonetka.cpp:156:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
     for (auto it : ans) cout << it << " "; cout << endl;
     ^~~
zagonetka.cpp:156:44: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
     for (auto it : ans) cout << it << " "; cout << endl;
                                            ^~~~
zagonetka.cpp:167:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
     for (auto it : ans) cout << it << " "; cout << endl;
     ^~~
zagonetka.cpp:167:44: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
     for (auto it : ans) cout << it << " "; cout << endl;
                                            ^~~~

• Subtask 1: 9 / 9
• Subtask 2: 18 / 18
• Subtask 3: 22 / 22
• Subtask 4: 51 / 51