Submission #1010224

#	Time	Username	Problem	Language	Result	Execution time	Memory
1010224		spacewalker	Dreaming (IOI13_dreaming)	C++17	100 / 100	66 ms	12272 KiB

dreaming

#include <bits/stdc++.h>

using namespace std;

#include "dreaming.h"

vector<pair<int, int>> distances(const vector<vector<pair<int, int>>> &graph, int start, vector<int> &dist, vector<int> &par) {
  queue<int> q;
  q.push(start);
  dist[start] = 0;
  vector<pair<int, int>> component;
  while (!q.empty()) {
    int cur = q.front(); q.pop();
    component.emplace_back(dist[cur], cur);
    for (auto [u, w] : graph[cur]) {
      if (dist[u] != -1) continue;
      dist[u] = dist[cur] + w;
      par[u] = cur;
      q.push(u);
    }
  }
  return component;
}

pair<int, int> find_cc_radius_diam(const vector<vector<pair<int, int>>> &graph, int start, vector<int> &dist, vector<int> &par) {
  vector<pair<int, int>> cc = distances(graph, start, dist, par);
  int v1 = max_element(cc.begin(), cc.end())->second;
  // reset the dist array, so we can run the second BFS
  for (auto [_, v] : cc) dist[v] = -1;
  cc = distances(graph, v1, dist, par);
  int v2 = max_element(cc.begin(), cc.end())->second;
  // The "center" of the tree must be on the v1-v2 path.
  int diameter = dist[v2];
  int rad = diameter;
  for (; v2 != v1; v2 = par[v2]) {
    rad = min(rad, max(dist[v2], diameter - dist[v2]));
  }
  return {rad, diameter};
}

int travelTime(int N, int M, int L, int A[], int B[], int T[]) {
  vector<int> radius;
  vector<vector<pair<int, int>>> graph(N);
  for (int i = 0; i < M; ++i) {
    graph[A[i]].emplace_back(B[i], T[i]);
    graph[B[i]].emplace_back(A[i], T[i]);
  }
  vector<int> visited(N, -1), par(N, -1);
  int ans = 0;
  for (int i = 0; i < N; ++i) {
    if (visited[i] != -1) continue;
    auto [r, d] = find_cc_radius_diam(graph, i, visited, par);
    radius.push_back(r);
    ans = max(ans, d);
  }
  sort(radius.begin(), radius.end(), greater());
  if (radius.size() == 1) {
    return ans;
  } else if (radius.size() == 2) {
    return max(ans, radius[0] + radius[1] + L);
  } else {
    return max(ans, max(radius[0] + radius[1] + L, radius[1] + radius[2] + 2 * L));
  }
}

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