Submission #918052

#	Time	Username	Problem	Language	Result	Execution time	Memory
918052		vjudge1	Easter Eggs (info1cup17_eastereggs)	C++17	100 / 100	11 ms	1252 KiB

eastereggs

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")
 
#include <bits/stdc++.h>
#include "grader.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
/* DEFINES */
#define S second
#define F first
#define ll long long
#define ull unsigned long long
#define ld long double
#define npos ULLONG_MAX
#define INF LLONG_MAX
#define vv(a) vector<a>
#define pp(a, b) pair<a, b>
#define pq(a) priority_queue<a>
#define qq(a) queue<a>
#define ss(a) set<a>
#define mm(a, b) map<a, b>
#define ump(a, b) unordered_map<a, b>
#define ANDROID                   \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define elif else if
#define endl "\n"
#define allc(a) begin(a), end(a)
#define all(a) a, a + sizeof(a) / sizeof(a[0])
#define pb push_back
#define logi(a) __lg(a)
#define sqrt(a) sqrtl(a)
#define mpr make_pair
#define ins insert
using namespace std;
using namespace __gnu_pbds;
using namespace __cxx11;
typedef char chr;
typedef basic_string<chr> str;
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll N = 515;
vv(ll) g[N];
vv(ll) order;
void dfs(ll v, ll par = 0)
{
    order.pb(v);
    for (ll to : g[v])
        if (to != par)
            dfs(to, v);
}
int findEgg(int n, vector<pair<int, int>> e)
{
    order.clear();
    for (ll i = 1; i <= n; i++)
        g[i].clear();
    for (ll i = 0; i < n - 1; i++)
        g[e[i].F].pb(e[i].S), g[e[i].S].pb(e[i].F);
    dfs(1);
    ll l = 0, r = n - 1;
    vv(int) v;
    while (l < r)
    {
        ll mid = (l + r) >> 1;
        v.clear();
        for (ll i = 0; i <= mid; i++)
            v.pb(order[i]);
        if (query(v))
            r = mid;
        else
            l = mid + 1;
    }
    return order[r];
}

• Subtask 1: 40 / 40
• Subtask 2: 40 / 40
• Subtask 3: 20 / 20