結果

問題 No.2231 Surprising Flash!
ユーザー 👑 AngrySadEight
提出日時 2023-02-14 03:24:38
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 4,083 bytes
コンパイル時間 3,841 ms
コンパイル使用メモリ 147,436 KB
最終ジャッジ日時 2025-02-10 15:13:40
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 41 WA * 3
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ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <string>
#include <vector>
#include <atcoder/string>
#include <atcoder/convolution>
#include <atcoder/segtree>
using namespace std;
using namespace atcoder;
typedef long long ll;
template<typename T> void debug(vector<T> &v){
    for (ll i = 0; i < v.size(); i++){
        cerr << v[i] << " ";
    }
    cerr << endl;
}
ll ctol(char c){
    ll ret;
    if (c == '?') ret = 0;
    else{
        ret = (ll)(c - 'a' + 1);
    }
    return ret;
}
vector<ll> wild_card_matching(string &S, string &T){
    ll len_S = S.size();
    ll len_T = T.size();
    vector<vector<ll>> pow_S(4, vector<ll>(len_S));
    vector<vector<ll>> pow_T(4, vector<ll>(len_T));
    for (ll i = 0; i < len_S; i++){
        pow_S[1][i] = ctol(S[i]);
        for (ll j = 1; j < 3; j++){
            pow_S[j + 1][i] = pow_S[j][i] * pow_S[1][i];
        }
    }
    for (ll i = 0; i < len_T; i++){
        pow_T[1][i] = ctol(T[i]);
        for (ll j = 1; j < 3; j++){
            pow_T[j + 1][i] = pow_T[j][i] * pow_T[1][i];
        }
    }
    for (ll i = 0; i < 4; i++){
        reverse(pow_T[i].begin(), pow_T[i].end());
    }
    vector<ll> vec_conv1 = convolution_ll(pow_S[3], pow_T[1]);
    vector<ll> vec_conv2 = convolution_ll(pow_S[2], pow_T[2]);
    vector<ll> vec_conv3 = convolution_ll(pow_S[1], pow_T[3]);
    vector<ll> ret(0);
    for (ll i = 0; i < len_S - len_T + 1; i++){
        if (vec_conv1[len_T - 1 + i] - 2 *  vec_conv2[len_T - 1 + i] + vec_conv3[len_T - 1 + i] == 0){
            ret.push_back(i);
        }
    }
    return ret;
}
ll op(ll a, ll b){
    return min(a, b);
}
ll e(){
    return 1e9;
}
int main(){
    ll T;
    cin >> T;
    while(T--){
        ll N, M;
        cin >> N >> M;
        string S1;
        cin >> S1;
        string S2;
        cin >> S2;
        vector<ll> wcm = wild_card_matching(S1, S2);
        if (wcm.size() == 0){
            cout << -1 << endl;
            continue;
        }
        //debug(wcm);
        string S3 = "";
        for (ll i = 0; i < N; i++){
            if (S1[i] == '?') S3.push_back('a');
            else S3.push_back(S1[i]);
        }
        //S3 + S2 の LCP を求める
        vector<int> sa = suffix_array(S3 + S2);
        vector<int> lcp = lcp_array(S3 + S2, sa);
        vector<int> rev_sa(N + M);
        for (int i = 0; i < N + M; i++){
            rev_sa[sa[i]] = i;
        }
        segtree<ll, op, e> seg(N + M);
        for (ll i = 0; i < N + M; i++){
            seg.set(i, lcp[i]);
        }
        ll idx = 0;
        for (ll i = 1; i < wcm.size(); i++){
            ll left, right;
            ll diffs = wcm[i] - wcm[idx];
            if (diffs >= M){
                idx = i;
                continue;
            }
            left = min(rev_sa[wcm[idx]], rev_sa[N]);
            right = max(rev_sa[wcm[idx]], rev_sa[N]);
            ll lcp1 = seg.prod(left, right);
            if (lcp1 < diffs){
                if (S3[wcm[i] + lcp1] < S2[lcp1]){
                    idx = i;
                    continue;
                }
                else{
                    continue;
                }
            }
            left = min(rev_sa[N], rev_sa[N + diffs]);
            right = max(rev_sa[N], rev_sa[N + diffs]);
            ll lcp2 = seg.prod(left, right);
            if (lcp2 < M - diffs){
                if (S2[lcp2] < S2[diffs + lcp2]){
                    idx = i;
                    continue;
                }
                else{
                    continue;
                }
            }
            left = min(rev_sa[wcm[idx] + M], rev_sa[N + M - diffs]);
            right = max(rev_sa[wcm[idx] + M], rev_sa[N + M - diffs]);
            ll lcp3 = seg.prod(left, right);
            if (lcp3 < diffs){
                if (S2[M - diffs + lcp3] < S3[wcm[idx] + M + lcp3]){
                    idx = i;
                    continue;
                }
                else{
                    continue;
                }
            }
        }
        cout << S3.substr(0, wcm[idx]) << S2 << S3.substr(wcm[idx] + M, N - wcm[idx]) << endl;
    }
}
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