結果

問題 No.2319 Friends+
コンテスト
ユーザー hliuser1
提出日時 2023-12-25 20:23:56
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 4,832 bytes
コンパイル時間 6,144 ms
コンパイル使用メモリ 257,996 KB
最終ジャッジ日時 2025-02-18 14:38:16
ジャッジサーバーID
（参考情報） 		judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 5 TLE * 1 -- * 39
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main2()’:
main.cpp:110:1: warning: no return statement in function returning non-void [-Wreturn-type]
  110 | }
      | ^

ソースコード

diff # 

// https://yukicoder.me/problems/no/2319
// really cool problem
//
// Solution 0. use bitsets and bruteforce
//
// Solution 1. sqrt over nodes
// let K = sqrt(M)
// a person is "few" if it has < K friends and "many" if >= K
// there are at most 2M / K ~= K people that are "many"
// to solve a query where x is few, brute force it
// to solve a query where x is many, consider its friends that are:
//     a. many - you can check ALL many's for the condition
//     b. few  - store this as map f[i][j] = # of few friends of the many i in world j
// update naively, it works in O(Q*sqrt M + N*sqrt M + M)
//
// Solution 2. sqrt over queries
// for each chunk, recompute a map of for each person i, how many friends they have in world j
// then for each query, loop over the previous queries in the chunk and simulate. if by the end,
// you still have at least one friend in world[y], ur good
// the map computations take M sqrt(Q) time total, the rest is Q sqrt(Q)
#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
using namespace std;

using HashSet = unordered_set<int>;
using FewCounter = unordered_map<int, vector<int>>;

signed main2() {
  cin.tie(0)->sync_with_stdio(0);
  int N, M, Q;
  cin >> N >> M;
  int K = sqrt(M);    
  vector<int>  world(N+1);  // world[i] = the world that person i is in
  FewCounter   num_few_of;  // num_few_of[i].at(w) = # friends of i in world j with < K friends
  vector<int>  many;        // many = [those with >= K friends]
  vector<HashSet> friends(N+1);

  auto is_many = [&] (int i) { 
    return friends[i].size() >= K;
  };

  for (int i = 1; i <= N; ++i)
    cin >> world[i];

  for (int i = 1; i <= M; ++i) {
    int a, b; 
    cin >> a >> b;
    friends[a].insert(b);
    friends[b].insert(a);
  }

  for (int i = 1; i <= N; ++i) {
    if (!is_many(i)) continue;
    many.push_back(i);
    num_few_of[i].assign(N+1, 0);
    for (int f : friends[i]) 
      if (!is_many(f)) {
        int w = world[f];
        ++num_few_of[i].at(w);
      }
  }

  auto solve_many = [&] (int x, int y) {
    int new_world = world[y];
    if (num_few_of[x].at(new_world) > 0)
      goto success;

    for (int m : many) 
      if (friends[x].count(m) && world[m] == new_world) 
        goto success;
      
    return "No\n";

    success:
    world[x] = new_world;
    return "Yes\n";
  };

  auto solve_few = [&] (int x, int y) {
    for (int f : friends[x]) 
      if (world[f] == world[y]) 
        goto success;

    return "No\n";

    success:
    int old_world = world[x];
    int new_world = world[y];
    for (int f : friends[x]) {
      if (!is_many(f)) continue;
      --num_few_of[f].at(old_world);
      ++num_few_of[f].at(new_world);
    }
    world[x] = new_world;
    return "Yes\n";
  };

  cin >> Q;
  while (Q--) {
    int x, y;
    cin >> x >> y;
    if (world[x] == world[y]) {
      cout << "No\n";
      continue;
    }
    auto ans = is_many(x) ? solve_many(x,y) : solve_few(x,y);
    cout << ans;
  }
}

// SOLUTION 2: TLEs sadly. could not get it to pass
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
constexpr size_t MXN = 20001;
constexpr size_t MXM = 400000;
// constexpr size_t MXN = 100;
// constexpr size_t MXM = 100;

signed main() {
  cin.tie(0)->sync_with_stdio(0);
  int N, M, Q;
  cin >> N >> M;
  
  short world[MXN];
  bitset<MXN> friends[MXN];
  short friendships[MXM];
  gp_hash_table<int,short> initial_count[MXN];
  // initial_count[i][j] = # of friends of i in world j

  for (int i = 1; i <= N; ++i) 
    cin >> world[i];

  for (int i = 0; i < M; ++i) {
    short a, b;
    cin >> a >> b;
    friendships[2*i] = a, friendships[2*i+1] = b;
    friends[a][b] = true;
    friends[b][a] = true;
  }

  auto reset_map = [&] {
    for (int i = 1; i <= N; ++i)
      initial_count[i].clear();
    for (int i = 0; i < M; ++i) {
      short a = friendships[2*i];
      short b = friendships[2*i+1];
      ++initial_count[a][world[b]];
      ++initial_count[b][world[a]];
    }
  };

  cin >> Q;
  int K = sqrt(Q);
  int q = 0;
  gp_hash_table<int,vector<int>> succeeded_wx, succeeded_wy;

  while (q < Q) {
    reset_map();
    for (int k = 0; k < K && q < Q; ++k, ++q) {
      short x, y;
      cin >> x >> y;
      if (world[x] == world[y]) {
        cout << "No\n";
        continue;
      }

      int start = initial_count[x][world[y]];
      for (int xl : succeeded_wx[ world[y] ]) {
        start -= (friends[x][xl]);
      }
      for (int xl : succeeded_wy[ world[y] ]) {
        start += (friends[x][xl]);
      }
      if (start > 0) {
        cout << "Yes\n";
        succeeded_wx[world[x]].push_back(x);
        succeeded_wy[world[y]].push_back(x);
        world[x] = world[y];
      }
      else cout << "No\n";
    }
    succeeded_wx.clear();
    succeeded_wy.clear();
  }
  return 0;
}
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