結果
問題 | No.2319 Friends+ |
ユーザー | hliuser1 |
提出日時 | 2023-12-25 18:39:41 |
言語 | C++17(clang) (17.0.6 + boost 1.83.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 4,655 bytes |
コンパイル時間 | 4,193 ms |
コンパイル使用メモリ | 167,808 KB |
実行使用メモリ | 13,768 KB |
最終ジャッジ日時 | 2024-09-27 14:35:01 |
合計ジャッジ時間 | 10,173 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
(要ログイン)
テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
13,756 KB |
testcase_01 | AC | 2 ms
6,812 KB |
testcase_02 | TLE | - |
testcase_03 | -- | - |
testcase_04 | -- | - |
testcase_05 | -- | - |
testcase_06 | -- | - |
testcase_07 | -- | - |
testcase_08 | -- | - |
testcase_09 | -- | - |
testcase_10 | -- | - |
testcase_11 | -- | - |
testcase_12 | -- | - |
testcase_13 | -- | - |
testcase_14 | -- | - |
testcase_15 | -- | - |
testcase_16 | -- | - |
testcase_17 | -- | - |
testcase_18 | -- | - |
testcase_19 | -- | - |
testcase_20 | -- | - |
testcase_21 | -- | - |
testcase_22 | -- | - |
testcase_23 | -- | - |
testcase_24 | -- | - |
testcase_25 | -- | - |
testcase_26 | -- | - |
testcase_27 | -- | - |
testcase_28 | -- | - |
testcase_29 | -- | - |
testcase_30 | -- | - |
testcase_31 | -- | - |
testcase_32 | -- | - |
testcase_33 | -- | - |
testcase_34 | -- | - |
testcase_35 | -- | - |
testcase_36 | -- | - |
testcase_37 | -- | - |
testcase_38 | -- | - |
testcase_39 | -- | - |
testcase_40 | -- | - |
testcase_41 | -- | - |
testcase_42 | -- | - |
testcase_43 | -- | - |
testcase_44 | -- | - |
testcase_45 | -- | - |
testcase_46 | -- | - |
ソースコード
// 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 #include <bits/stdc++.h> #pragma GCC optimize("O3") #pragma GCC target("sse4") using namespace std; using HashSet = unordered_set<int>; using FewCounter = unordered_map<int, vector<int>>; // signed main2() { // cout << "sol 1\n"; // freopen("/Users/drymerge/CLionProjects/DEBUG/a.in", "r", stdin); // freopen("/Users/drymerge/a.out", "w", stdout); // 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; // } // } signed main() { cin.tie(0)->sync_with_stdio(0); // cout << "sol2\n"; // freopen("/Users/drymerge/CLionProjects/DEBUG/a.in", "r", stdin); // freopen("/Users/drymerge/b.out", "w", stdout); int N, M, Q; cin >> N >> M; vector<int> world(N+1); for (int i = 1; i <= N; ++i) cin >> world[i]; vector<unordered_set<int>> friends(N+1); vector<array<int,2>> friendships(M); unordered_map<int, unordered_map<int, int>> initial_count; // initial_count[i][j] = # of friends of i in world j for (auto& [a,b] : friendships) { cin >> a >> b; friends[a].insert(b); friends[b].insert(a); } auto reset_map = [&] { initial_count.clear(); for (auto [a,b] : friendships) { ++initial_count[a][world[b]]; ++initial_count[b][world[a]]; } }; auto is_friends = [&] (int i, int j) -> bool { if (friends[i].size() > friends[j].size()) swap(i, j); return friends[i].count(j); }; cin >> Q; int K = sqrt(Q); int q = 0; vector<array<int,4>> succeeded; while (q < Q) { reset_map(); for (int k = 0; k < K && q < Q; ++k, ++q) { int x, y; cin >> x >> y; if (world[x] == world[y]) { cout << "No\n"; continue; } int start = initial_count[x][world[y]]; for (auto [xl,yl,wx,wy] : succeeded) { if (wx == world[y] && is_friends(x, xl)) --start; if (wy == world[y] && is_friends(x, xl)) ++start; } if (start > 0) { cout << "Yes\n"; succeeded.push_back( {x,y,world[x],world[y]} ); world[x] = world[y]; } else cout << "No\n"; } succeeded.clear(); } return 0; }