Submission #1274835

#	Time	Username	Problem	Language	Result	Execution time	Memory
1274835		mehrzad	이주 (IOI25_migrations)	C++17	10 / 100	35 ms	1028 KiB

migrations

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

/*
  General solution (works for both subtasks):

  During the first run (send_message calls in increasing i):
  - We build the tree online and maintain binary-lifting parents (up[k][i]) and depths.
  - We maintain the current diameter endpoints (A, B) and its length D.
    For each new node i with parent p:
      * Compute dist(i, A) and dist(i, B) using LCA.
      * If dist(i, A) > D, set (B = i) and update D.
      * else if dist(i, B) > D, set (A = i) and update D.
  - We send exactly TWO messages total:
      * at i == N-2 we send A+1
      * at i == N-1 we send B+1
    (Each is ≤ N ≤ 10000 ≤ 20000, and M = 2 ≤ 50.)

  During the second run (longest_path):
  - Read the last two non-zero entries of S (order matters):
      * the penultimate non-zero is A+1, the last non-zero is B+1.
  - Return (A, B).

  This yields the true diameter for any tree (no need for the "root is an endpoint" assumption).
*/

// ---------------- Persistent state for the first run ----------------
namespace {
  constexpr int MAXN = 10000 + 5;
  constexpr int LOG   = 15; // since 2^14 = 16384 > 10000

  static bool initialized = false;
  static int  depth_arr[MAXN];
  static int  up[LOG][MAXN]; // up[k][v] = 2^k-th ancestor of v, or -1
  static int  A = 0, B = 0;  // current diameter endpoints
  static int  D = 0;         // current diameter length

  inline int lca(int u, int v) {
    if (u == -1 || v == -1) return u == -1 ? v : u;
    if (depth_arr[u] < depth_arr[v]) swap(u, v);
    int diff = depth_arr[u] - depth_arr[v];
    for (int k = 0; k < LOG; ++k) if (diff & (1 << k)) u = up[k][u];
    if (u == v) return u;
    for (int k = LOG - 1; k >= 0; --k) {
      if (up[k][u] != up[k][v]) {
        u = up[k][u];
        v = up[k][v];
      }
    }
    return up[0][u];
  }

  inline int dist(int u, int v) {
    int w = lca(u, v);
    return depth_arr[u] + depth_arr[v] - 2 * depth_arr[w];
  }

  inline void add_node(int i, int p) {
    // Set depth and ancestors for node i
    depth_arr[i] = depth_arr[p] + 1;
    up[0][i] = p;
    for (int k = 1; k < LOG; ++k) {
      int mid = up[k - 1][i];
      up[k][i] = (mid == -1 ? -1 : up[k - 1][mid]);
    }

    // Try to improve diameter with endpoint A or B
    int dA = dist(i, A);
    if (dA > D) {
      B = i;
      D = dA;
      return;
    }
    int dB = dist(i, B);
    if (dB > D) {
      A = i;
      D = dB;
      return;
    }
  }
}

// ---------------- Research team procedure (first run) ----------------
int send_message(int N, int i, int Pi) {
  if (!initialized) {
    initialized = true;

    // Initialize root (node 0)
    depth_arr[0] = 0;
    for (int k = 0; k < LOG; ++k) up[k][0] = -1;

    // Start diameter as (0,0)
    A = 0; B = 0; D = 0;
  }

  // Add node i with parent Pi
  add_node(i, Pi);

  // Send two messages only at the last two sites
  if (i == N - 2) {
    return A + 1; // encode A
  }
  if (i == N - 1) {
    return B + 1; // encode B
  }
  return 0;
}

// ---------------- Museum procedure (second run) ----------------
std::pair<int,int> longest_path(std::vector<int> S) {
  // Extract the last two non-zero messages: penultimate -> A+1, last -> B+1
  int N = (int)S.size();
  int last_idx = -1, penult_idx = -1;
  for (int i = 1; i < N; ++i) {
    if (S[i] != 0) {
      penult_idx = last_idx;
      last_idx = i;
    }
  }

  if (last_idx == -1) {
    // No messages found; fallback to trivial answer
    return {0, 0};
  }
  if (penult_idx == -1) {
    // Only one message found; interpret it as B+1 with A assumed 0 as a best-effort fallback
    int b = S[last_idx] - 1;
    if (b < 0) b = 0;
    return {0, b};
  }

  int a = S[penult_idx] - 1;
  int b = S[last_idx] - 1;
  if (a < 0) a = 0;
  if (b < 0) b = 0;
  return {a, b};
}

• Subtask 0: 0 / 0
• Subtask 1: 10 / 30
• Subtask 2: 0 / 70