Submission #1058102

#	Time	Username	Problem	Language	Result	Execution time	Memory
1058102		caterpillow	Fun Tour (APIO20_fun)	C++17	21 / 100	52 ms	19088 KiB

fun

#include <bits/stdc++.h>

using namespace std;

using db = long double;
using ll = long long;
using pl = pair<ll, ll>;
using pi = pair<int, int>;
#define vt vector
#define f first
#define s second
#define pb push_back
#define all(x) x.begin(), x.end() 
#define size(x) ((int) (x).size())
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define F0R(i, b) FOR (i, 0, b)
#define endl '\n'
const ll INF = 1e18;
const int inf = 1e9;

template<typename... Args> // tuples
ostream& operator<<(ostream& os, tuple<Args...> t) { 
    apply([&](Args... args) { string dlm = "{"; ((os << dlm << args, dlm = ", "), ...); }, t);
    return os << "}";
}

template<typename T, typename V> // pairs
ostream& operator<<(ostream& os, pair<T, V> p) { return os << "{" << p.f << ", " << p.s << "}"; }

template<class T, class = decltype(begin(declval<T>()))> // iterables
typename enable_if<!is_same<T, string>::value, ostream&>::type operator<<(ostream& os, const T& v) {
    string dlm = "{";
    for (auto& i : v) os << dlm << i, dlm = ", ";
    return os << "}";  
}

template <typename T, typename... V>
void printer(string pfx, const char *names, T&& head, V&& ...tail) {
    int i = 0;
    while (names[i] && names[i] != ',') i++;
    constexpr bool is_str = is_same_v<decay_t<T>, const char*>;
    if (is_str) cerr << " " << head;
    else cerr << pfx, cerr.write(names, i) << " = " << head; 
    if constexpr (sizeof...(tail)) printer(is_str ? "" : ",", names + i + 1, tail...);
    else cerr << endl;
}

#ifdef LOCAL
#define dbg(...) printer(to_string(__LINE__) + ": ", #__VA_ARGS__, __VA_ARGS__)
#else
#define dbg(x...)
#define cerr if (0) std::cerr
#endif

#include "fun.h"

/*

we can query distance and rooted subtree size

subtask 1 and 2: n^2

subtask 3: perfect binary tree, rooted at 0 with segtree indexing (kinda)

subtask 4: im pretty sure this is meant to be a star graph

repeatedly taking the diameter will give a valid tour trust

*/

int n, q;

vt<int> createFunTour(int N, int Q) {
    n = N;
    q = Q;
    vt<int> out;

    n++;
    vt<int> used(n);
    vt<pl> dp(2 * n);
    vt<int> depth(n);
    depth[1] = 0;
    FOR (i, 2, n) depth[i] = depth[i / 2] + 1;

    auto pull = [&] (int u) {
        dp[u] = max({{depth[u], u}, dp[2 * u], dp[2 * u + 1]});
    };

    ROF (i, 1, n) pull(i);

    auto die = [&] (int u) {
        dp[u] = {};
        while (u /= 2) pull(u);
    };

    dbg(depth);
    dbg(dp);

    int u = n - 1;
    used[n - 1] = used[0] = true;
    out.pb(u - 1);
    die(u);

    F0R (i, n - 2) {
        int v = u;
        pi best = max(dp[2 * u], dp[2 * u + 1]); // {dist, node with furthest dist}
        while (true) {
            if (used[v / 2]) break;
            best = max(best, {depth[u] - depth[v / 2], v / 2});
            best = max(best, {depth[u] - depth[v / 2] * 2 + dp[v ^ 1].f, dp[v ^ 1].s});
            v /= 2;
        }
        cerr << "furthest node from " << u << " is " << best.s << endl;
        used[best.s] = true;
        out.pb(best.s - 1);
        die(best.s);
        u = best.s;
    }

    dbg(out);
    return out;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 16
• Subtask 3: 21 / 21
• Subtask 4: 0 / 19
• Subtask 5: 0 / 34