結果

問題 No.2319 Friends+
ユーザー hliuser1hliuser1
提出日時 2023-12-26 06:07:09
言語 C++23
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 2,377 ms / 3,000 ms
コード長 6,280 bytes
コンパイル時間 5,042 ms
コンパイル使用メモリ 307,736 KB
実行使用メモリ 72,040 KB
最終ジャッジ日時 2024-09-27 14:56:39
合計ジャッジ時間 32,338 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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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 19 ms
57,008 KB
testcase_01 AC 19 ms
57,056 KB
testcase_02 AC 800 ms
69,048 KB
testcase_03 AC 745 ms
69,504 KB
testcase_04 AC 635 ms
69,248 KB
testcase_05 AC 716 ms
69,120 KB
testcase_06 AC 628 ms
69,036 KB
testcase_07 AC 827 ms
71,576 KB
testcase_08 AC 805 ms
71,656 KB
testcase_09 AC 792 ms
71,588 KB
testcase_10 AC 794 ms
71,604 KB
testcase_11 AC 770 ms
71,652 KB
testcase_12 AC 20 ms
57,184 KB
testcase_13 AC 20 ms
57,192 KB
testcase_14 AC 20 ms
57,140 KB
testcase_15 AC 21 ms
57,192 KB
testcase_16 AC 21 ms
57,192 KB
testcase_17 AC 19 ms
57,012 KB
testcase_18 AC 230 ms
58,984 KB
testcase_19 AC 629 ms
71,680 KB
testcase_20 AC 437 ms
68,708 KB
testcase_21 AC 387 ms
65,536 KB
testcase_22 AC 471 ms
69,684 KB
testcase_23 AC 636 ms
71,168 KB
testcase_24 AC 570 ms
71,012 KB
testcase_25 AC 532 ms
71,728 KB
testcase_26 AC 532 ms
71,272 KB
testcase_27 AC 528 ms
70,912 KB
testcase_28 AC 522 ms
70,832 KB
testcase_29 AC 486 ms
70,380 KB
testcase_30 AC 453 ms
69,556 KB
testcase_31 AC 416 ms
68,096 KB
testcase_32 AC 366 ms
64,868 KB
testcase_33 AC 318 ms
63,976 KB
testcase_34 AC 256 ms
60,780 KB
testcase_35 AC 241 ms
58,880 KB
testcase_36 AC 306 ms
72,040 KB
testcase_37 AC 1,390 ms
63,488 KB
testcase_38 AC 214 ms
58,720 KB
testcase_39 AC 189 ms
58,856 KB
testcase_40 AC 187 ms
58,852 KB
testcase_41 AC 184 ms
58,724 KB
testcase_42 AC 204 ms
58,948 KB
testcase_43 AC 267 ms
59,140 KB
testcase_44 AC 473 ms
60,264 KB
testcase_45 AC 1,367 ms
65,024 KB
testcase_46 AC 2,377 ms
70,580 KB
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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;

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;
template <typename K, typename V>
using HashTable = gp_hash_table<K,V>;

// template <typename K, typename V>
// using HashTable = unordered_map<K,V>;
constexpr size_t MXN = 20001;
constexpr size_t MXM = 400000;
// constexpr size_t MXN = 100;
// constexpr size_t MXM = 100;
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];
    // short initial_count[MXN][SQR];
    short X[SQR], Y[SQR];

    vector<int> whos_involved;

    // 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;
        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]];
            }
        }

        // 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;
    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;
}
0