Submission #1285959

#TimeUsernameProblemLanguageResultExecution timeMemory
1285959kawhietWorld Map (IOI25_worldmap)C++20
72 / 100
79 ms9564 KiB
#include <bits/stdc++.h>
#include "worldmap.h"
using namespace std;

vector<bool> vis;
vector<int> par, k;
vector<set<int>> t;
vector<vector<int>> g;

void dfs(int u) {
  vis[u] = 1;
  k.push_back(u);
  for (auto v : g[u]) {
    if (!vis[v]) {
      par[v] = u;
      dfs(v);
    }
  }
}

struct DSU {
  int n;
  vector<int> p;

  DSU(int sz) {
    n = sz;
    p.resize(n, -1);
  }

  int root(int x) {
    if (p[x] == -1) {
      return x;
    }
    return p[x] = root(p[x]);
  }

  bool link(int u, int v) {
    u = root(u);
    v = root(v);
    if (u == v) {
      return false;
    }
    p[u] = v;
    return true;
  }
};

vector<vector<int>> create_map(int n, int m, vector<int> a, vector<int> b) {
  k.clear();
  g.assign(n + 1, {});
  t.assign(n + 1, {});
  vis.assign(n + 1, 0);
  par.assign(n + 1, -1);
  DSU d(n + 1);
  for (int i = 0; i < m; i++) {
    if (d.link(a[i], b[i])) {
      g[a[i]].push_back(b[i]);
      g[b[i]].push_back(a[i]);
    } else {
      t[a[i]].insert(b[i]);
      t[b[i]].insert(a[i]);
    }
  }
  dfs(1);
  vector<int> res = {1};
  for (int i = 1; i < n; i++) {
    int u = res.back(), v = k[i];
    if (par[v] != u) {
      u = par[u];
      res.push_back(u);
      while (u != par[v]) {
        u = par[u];
        res.push_back(u);
      }
    }
    res.push_back(v);
  }
  int N = 4 * n, idx = 0;
  vector<vector<int>> ans(N, vector<int>(N));
  for (int i = 0; i < N; i++) {
    if (idx >= res.size()) {
      fill(ans[i].begin(), ans[i].end(), res.back());
      continue;
    }
    int u = res[idx++];
    fill(ans[i].begin(), ans[i].end(), u);
    if (t[u].empty()) {
      continue;
    }
    fill(ans[i + 2].begin(), ans[i + 2].end(), u);
    int j = 0;
    vector<int> rem;
    for (auto v : t[u]) {
      if (j + 2 >= N) break;
      ans[i + 1][j++] = u;
      ans[i + 1][j++] = v;
      rem.push_back(v);
    }
    for (auto x : rem) {
      t[u].erase(x);
      t[x].erase(u);
    }
    for (; j < N; j++) {
      ans[i + 1][j] = u;
    }
    i += 2;
  }
  return ans;
}
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