結果
| 問題 | No.2325 Skill Tree |
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2023-05-28 14:50:52 |
| 言語 | C++17 (gcc 15.2.0 + boost 1.90.0) |
| 結果 |
AC
|
| 実行時間 | 220 ms / 3,000 ms |
| コード長 | 1,115 bytes |
| 記録 | |
| コンパイル時間 | 3,059 ms |
| コンパイル使用メモリ | 277,052 KB |
| 実行使用メモリ | 14,464 KB |
| 最終ジャッジ日時 | 2026-06-30 22:49:26 |
| 合計ジャッジ時間 | 12,391 ms |
|
ジャッジサーバーID (参考情報) |
judge2_0 / judge1_0 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 36 |
コンパイルメッセージ
main.cpp: In function 'int main()':
main.cpp:23:36: warning: ignoring return value of 'std::basic_ostream<_CharT, _Traits>* std::basic_ios<_CharT, _Traits>::tie() const [with _CharT = char; _Traits = std::char_traits<char>]', declared with attribute 'nodiscard' [-Wunused-result]
23 | ios::sync_with_stdio(0);cin.tie();
| ~~~~~~~^~
In file included from /home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/ios:48,
from /home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/istream:42,
from /home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/sstream:42,
from /home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/complex:50,
from /home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/x86_64-pc-linux-gnu/bits/stdc++.h:141,
from main.cpp:1:
/home/linuxbrew/.linuxbrew/Cellar/gcc/15.2.0_1/include/c++/15/bits/basic_ios.h:310:7: note: declared here
310 | tie() const
| ^~~
ソースコード
#include <bits/stdc++.h>
#include <atcoder/all>
using ll = long long;
#define MOD 1000000007
#define Mod 998244353
const int MAX = 1000000005;
const long long INF = 1000000000000000005LL;
using namespace std;
using namespace atcoder;
vector<int> need(200005, MAX), L(200005);
void dfs(int v, int p, vector<vector<int>> &G, int mx = 1) {
mx = max(mx, L[v]);
need[v] = min(need[v], mx);
for (int nv : G[v]) {
if (nv == p) continue;
dfs(nv, v, G, mx);
}
}
int main() {
ios::sync_with_stdio(0);cin.tie();
int N;
cin >> N;
vector<vector<int>> G(N);
L[0] = 1;
for (int i = 1; i < N; i++) {
int a;
cin >> L[i] >> a; a--;
G[a].push_back(i);
}
dfs(0, -1, G);
int Q;
cin >> Q;
auto sorted = need;
sort(sorted.begin(), sorted.end());
while (Q--) {
int t, x;
cin >> t >> x;
if (t == 1) {
cout << upper_bound(sorted.begin(), sorted.end(), x) - sorted.begin() << endl;
} else {
x--;
cout << (need[x] == MAX ? -1 : need[x]) << endl;
}
}
}