結果
| 問題 |
No.2319 Friends+
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2023-12-25 18:44:16 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 1,912 ms / 3,000 ms |
| コード長 | 4,473 bytes |
| コンパイル時間 | 4,029 ms |
| コンパイル使用メモリ | 277,984 KB |
| 実行使用メモリ | 72,064 KB |
| 最終ジャッジ日時 | 2024-09-27 14:35:36 |
| 合計ジャッジ時間 | 34,133 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / 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
//
// 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 main() {
cin.tie(0)->sync_with_stdio(0);
// 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 main2() {
cout << "sol2\n";
freopen("/Users/drymerge/CLionProjects/DEBUG/a.in", "r", stdin);
freopen("/Users/drymerge/b.out", "w", stdout);
cin.tie(0)->sync_with_stdio(0);
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;
}