結果
| 問題 | No.2319 Friends+ |
| ユーザー |
 hliuser1
|
| 提出日時 | 2023-12-25 20:23:56 |
| 言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 4,832 bytes |
| コンパイル時間 | 5,268 ms |
| コンパイル使用メモリ | 266,316 KB |
| 実行使用メモリ | 78,184 KB |
| 最終ジャッジ日時 | 2023-12-25 20:24:16 |
| 合計ジャッジ時間 | 18,981 ms |
|
ジャッジサーバーID （参考情報） |
judge12 / judge13 |
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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 | 18 ms
63,844 KB |
| testcase_01 | AC | 19 ms
57,040 KB |
| testcase_02 | AC | 1,587 ms
72,916 KB |
| testcase_03 | AC | 1,545 ms
72,916 KB |
| testcase_04 | AC | 1,351 ms
72,916 KB |
| testcase_05 | AC | 1,521 ms
72,916 KB |
| testcase_06 | AC | 1,365 ms
72,916 KB |
| testcase_07 | TLE | - |
| 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 | -- | - |
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
// 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;
}