結果
| 問題 |
No.2319 Friends+
|
| コンテスト | |
| ユーザー |
 hliuser1
|
| 提出日時 | 2023-12-26 06:44:45 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 890 ms / 3,000 ms |
| コード長 | 5,836 bytes |
| コンパイル時間 | 3,596 ms |
| コンパイル使用メモリ | 297,988 KB |
| 実行使用メモリ | 102,656 KB |
| 最終ジャッジ日時 | 2024-09-27 15:00:13 |
| 合計ジャッジ時間 | 21,784 ms |
|
ジャッジサーバーID （参考情報） |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 45 |
ソースコード
// https://yukicoder.me/problems/no/2319
// really cool problem
//
// Solution 0. use bitsets and bruteforce (cheating)
//
// 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. ft. Jerry -> 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;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename K, typename V>
using HashTable = gp_hash_table<K,V>;
using HashSet = unordered_set<int>;
using FewCounter = HashTable<int, vector<int>>;
signed main() {
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: took forever to make it pass :/
// constexpr size_t MXN = 20001;
// constexpr size_t MXM = 400000;
// constexpr size_t SQR = 450;
// #define all(x) begin(x), end(x)
// signed main() {
// cin.tie(0)->sync_with_stdio(0);
// int N, M, Q;
// cin >> N >> M;
// short world[MXN];
// bitset<MXN> friends[MXN];
// vector<short> friendss[MXN];
// short friendships[MXM];
// HashTable<int, short> initial_count[MXN];
// // initial_count[i][j] = # of friends of i in world j
// short X[SQR], Y[SQR];
// vector<int> whos_involved;
// for (int i = 1; i <= N; ++i)
// cin >> world[i];
// for (int i = 0; i < M; ++i) {
// short a, b;
// cin >> a >> b;
// friendss[a].push_back(b);
// friendss[b].push_back(a);
// friendships[2*i] = a, friendships[2*i+1] = b;
// friends[a][b] = true;
// friends[b][a] = true;
// }
// auto reset_map = [&] {
// sort( all(whos_involved) );
// whos_involved.resize(unique(all(whos_involved)) - begin(whos_involved));
// for (int c : whos_involved) {
// if (!initial_count[c].empty())
// initial_count[c].clear();
// for (int f : friendss[c]) {
// ++initial_count[c][world[f]];
// }
// }
// };
// cin >> Q;
// int K = sqrt(Q);
// int q = 0;
// HashTable<int,vector<int>> succeeded_wx, succeeded_wy;
// while (q < Q) {
// int last = min(Q, q + K);
// for (int k = q; k < last; ++k) {
// cin >> X[k-q] >> Y[k-q];
// whos_involved.push_back(X[k-q]);
// whos_involved.push_back(Y[k-q]);
// }
// reset_map();
// for (int k = 0; k < K && q < Q; ++k, ++q) {
// short x = X[k], y = Y[k];
// 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();
// whos_involved.clear();
// }
// return 0;
// }