結果

問題 No.2319 Friends+
ユーザー hliuser1
提出日時 2023-12-26 06:17:02
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,776 ms / 3,000 ms
コード長 3,657 bytes
コンパイル時間 5,851 ms
コンパイル使用メモリ 309,476 KB
実行使用メモリ 71,984 KB
最終ジャッジ日時 2024-09-27 14:58:55
合計ジャッジ時間 34,585 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
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ファイルパターン 結果
sample AC * 2
other AC * 45
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ソースコード

diff # 

// 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>;

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;
}
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