結果

問題 No.874 正規表現間距離
ユーザー hitonanodehitonanode
提出日時 2019-08-30 22:22:58
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 4,876 bytes
コンパイル時間 1,644 ms
コンパイル使用メモリ 175,056 KB
実行使用メモリ 18,944 KB
最終ジャッジ日時 2024-11-22 00:27:44
合計ジャッジ時間 3,149 ms
ジャッジサーバーID
(参考情報)
judge3 / judge1
このコードへのチャレンジ
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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 2 ms
5,248 KB
testcase_01 AC 1 ms
5,248 KB
testcase_02 AC 2 ms
5,248 KB
testcase_03 AC 2 ms
5,248 KB
testcase_04 AC 2 ms
5,248 KB
testcase_05 AC 2 ms
5,248 KB
testcase_06 WA -
testcase_07 AC 2 ms
5,248 KB
testcase_08 AC 2 ms
5,248 KB
testcase_09 AC 2 ms
5,248 KB
testcase_10 WA -
testcase_11 AC 2 ms
5,248 KB
testcase_12 AC 2 ms
5,248 KB
testcase_13 AC 1 ms
5,248 KB
testcase_14 AC 2 ms
5,248 KB
testcase_15 AC 2 ms
5,248 KB
testcase_16 AC 5 ms
5,248 KB
testcase_17 AC 4 ms
5,248 KB
testcase_18 AC 2 ms
5,248 KB
testcase_19 AC 2 ms
5,248 KB
testcase_20 AC 2 ms
5,248 KB
testcase_21 AC 2 ms
5,248 KB
testcase_22 AC 2 ms
5,248 KB
testcase_23 AC 2 ms
5,248 KB
testcase_24 AC 2 ms
5,248 KB
testcase_25 AC 13 ms
8,576 KB
testcase_26 AC 12 ms
8,576 KB
testcase_27 AC 38 ms
18,944 KB
testcase_28 WA -
testcase_29 AC 45 ms
18,944 KB
testcase_30 WA -
testcase_31 WA -
testcase_32 WA -
testcase_33 AC 2 ms
5,248 KB
testcase_34 WA -
testcase_35 WA -
testcase_36 WA -
testcase_37 WA -
testcase_38 AC 2 ms
5,248 KB
testcase_39 AC 2 ms
5,248 KB
testcase_40 AC 1 ms
5,248 KB
testcase_41 AC 2 ms
5,248 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
///// This part below is only for debug, not used /////
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
///// END /////
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
using namespace __gnu_pbds; // find_by_order(), order_of_key()
template<typename TK> using pbds_set = tree<TK, null_type, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename TK, typename TV> using pbds_map = tree<TK, TV, less<TK>, rb_tree_tag, tree_order_statistics_node_update>;
*/

vector<int> seikei(string S)
{
    int N = S.length();
    vector<int> ret;

    REP(i, N)
    {
        if (i < N - 1)
        {
            if (S[i + 1] == '*') ret.emplace_back(S[i] + 100), i++;
            else if (S[i + 1] == '?') ret.emplace_back(S[i] + 2000), i++;
            else ret.emplace_back(S[i]);
        }
        else ret.emplace_back(S[i]);
    }
    return ret;
}

bool erasable(int c)
{
    return c >= 'a' + 60;
}
int main()
{
    string AS, BS;
    cin >> AS >> BS;
    vector<int> A = seikei(AS);
    vector<int> B = seikei(BS);
    vector<vector<int>> dp;
    int N = A.size(), M = B.size();

    ndarray(dp, N + 1, M + 1);
    REP(i, N + 1)
    {
        REP(j, M + 1)
        {
            dp[i][j] = 1e7;
        }
    }
    dp[0][0] = 0;
    REP(i, N + 1)
    {
        REP(j, M + 1)
        {
            if (i < N)
            {
                mmin(dp[i + 1][j], dp[i][j] + !erasable(A[i]));
            }
            if (j < M)
            {
                mmin(dp[i][j + 1], dp[i][j] + !erasable(B[j]));
            }
            if (i < N and j < M)
            {
                mmin(dp[i + 1][j + 1], dp[i][j] + abs(A[i] - B[j]) % 100);
                if (A[i] - B[j] == -100) mmin(dp[i + 1][j], dp[i][j]);
                if (A[i] - B[j] == 100) mmin(dp[i][j + 1], dp[i][j]);
            }
        }
    }
    cout << dp[N][M] << endl;
}
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