Submission #740295

#	Time	Username	Problem	Language	Result	Execution time	Memory
740295		Nelt	Dreaming (IOI13_dreaming)	C++17	100 / 100	134 ms	20156 KiB

dreaming

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")

#include "dreaming.h"
#include <bits/stdc++.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 sync                      \
    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 = 1e5 + 5;
vv(pp(ll, ll)) g[N];
ll depth[3][N];
bool used[N];
vv(ll) comp;
void dfs(ll v, ll t)
{
    used[v] = 1;
    if (!t)
        comp.pb(v);
    for (auto [to, w] : g[v])
        if (!used[to])
            depth[t][to] = depth[t][v] + w, dfs(to, t);
}
int travelTime(int n, int m, int cost, int A[], int B[], int C[]) {
    for (ll i = 0; i < m; i++)
        A[i]++, B[i]++, g[A[i]].pb(mpr(B[i], C[i])), g[B[i]].pb(mpr(A[i], C[i]));
    ll a, b;
    multiset<ll> s;
    ll ans = 0;
    for (ll i = 1; i <= n; i++)
        if (!used[i])
        {
            comp.clear();
            dfs(i, 0);
            a = i;
            for (ll j : comp)
                if (depth[0][j] > depth[0][a])
                    a = j;
            for (ll j : comp)   used[j] = 0;
            dfs(a, 1);
            b = a;
            for (ll j : comp)
                if (depth[1][j] > depth[1][b])
                    b = j;
            for (ll j : comp)   used[j] = 0;
            dfs(b, 2);
            ans = max(ans, depth[1][b]);
            ll mn = INF;
            for (ll j : comp)
                mn = min(mn, max(depth[1][j], depth[2][j]));
            s.ins(mn);
            // cout << mn << " ";
        }
    while (s.size() > 1)
    {
        ll x = *s.begin(), y = *s.rbegin();
        s.erase(s.begin());
        s.erase(--s.end());
        ans = max(ans, y + cost + x);
        s.ins(max(x + cost, y));
    }
    return max(*s.begin(), ans);
}

• Subtask 1: 14 / 14
• Subtask 2: 10 / 10
• Subtask 3: 23 / 23
• Subtask 4: 18 / 18
• Subtask 5: 12 / 12
• Subtask 6: 23 / 23