結果
| 問題 | No.3427 Erasing a Subsequence |
| コンテスト | |
| ユーザー |
Ohuton_Racing
|
| 提出日時 | 2026-01-11 14:55:40 |
| 言語 | C++17 (gcc 15.2.0 + boost 1.89.0) |
| 結果 |
MLE
|
| 実行時間 | - |
| コード長 | 2,741 bytes |
| 記録 | |
| コンパイル時間 | 4,364 ms |
| コンパイル使用メモリ | 281,484 KB |
| 実行使用メモリ | 814,592 KB |
| 最終ジャッジ日時 | 2026-01-11 14:55:47 |
| 合計ジャッジ時間 | 6,348 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 5 MLE * 1 -- * 8 |
ソースコード
#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
int main(){
int n, m;
cin >> n >> m;
vector<int> s(n), t(m);
cin >> s >> t;
vector<vector<vector<int>>> memo(n + 1, vector<vector<int>>(m + 1));
vector<int> NG = {1001001001};
rep(i, n + 1) rep(j, m + 1) memo[i][j] = NG;
memo[0][0] = vector<int>(0);
auto update = [&](int i, int key, auto& vec){
chmin(memo[i][key], vec);
};
rep(i, n){
rep(key, m + 1){
auto& vec = memo[i][key];
if(vec == NG) continue;
if(key < m and t[key] == s[i]){
update(i + 1, key + 1, vec);
}
vec.push_back(s[i]);
update(i + 1, key, vec);
vec.pop_back();
}
}
cout << memo[n][m];
return 0;
}
Ohuton_Racing