Submission #1242367

#	Time	Username	Problem	Language	Result	Execution time	Memory
1242367		AMel0n	Dreaming (IOI13_dreaming)	C++20	100 / 100	68 ms	22840 KiB

dreaming

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

typedef long long ll;
#define FOR(i,N) for(ll i = 0; i < N; i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second

#include "dreaming.h"

ll N, M, L;
vector<vector<pair<ll,ll>>> adj;
vector<signed char> vis;

pair<ll,ll> mxDist(ll n, ll p, ll d) {
    vis[n] = 1;
    pair<ll,ll> ans = {d, n}; // furthest dist, furthest node
    for(pair<ll,ll> c: adj[n]) {
        if (c.F == p) continue;
        pair<ll,ll> ret = mxDist(c.F, n, d + c.S);
        if (ret.F > ans.F) ans = ret;
    }
    return ans;
}

signed char rad(ll n, ll p, ll d, ll e, unordered_set<ll> &dist) {
    if (n == e) return 1;
    signed char found = 0;
    for(pair<ll,ll> c: adj[n]) {
        if (c.F == p) continue;
        if (rad(c.F, n, d + c.S, e, dist) == 1) found = 1;
    }
    if (found) dist.insert(d);
    return found;
}

int travelTime(int N, int M, int L, int A[], int B[], int T[]) {
    ::N = N, ::M = M, ::L = L;
    adj.resize(N);
    vis.resize(N);
    FOR(i, M) {
        adj[A[i]].push_back({B[i], T[i]});
        adj[B[i]].push_back({A[i], T[i]});
    }

    ll res = 0;
    vector<ll> radii; 
    // let optimal node = node in a tree that best splits the diameter in half
    // let radii = maximum distance from any node in a tree to optimal node 
    FOR(i, N) {
        if (vis[i]) continue;
        ll startp = mxDist(i, -1, 0).S;
        auto [diam, endp] = mxDist(startp, -1, 0);
        res = max(res, diam);
        unordered_set<ll> dist;
        rad(startp, -1, 0, endp, dist);
        ll minrad = 1e18;
        for(ll d: dist) minrad = min(minrad, max(d, diam - d));
        if (minrad == 1e18) radii.push_back(0);
        else radii.push_back(minrad);
    }
    sort(all(radii));
    reverse(all(radii));
    // connect tree with the largest radii to all other trees (via respective optimal nodes)
    if ((ll)radii.size() >= 2) res = max(radii[0] + radii[1] + L, res);
    if ((ll)radii.size() >= 3) res = max(radii[1] + radii[2] + L*2, res);
    return (int)res;
}

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