結果

問題 No.422 文字列変更 (Hard)
ユーザー fumiphys
提出日時 2019-07-27 17:24:41
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 23 ms / 3,000 ms
コード長 5,210 bytes
コンパイル時間 1,553 ms
コンパイル使用メモリ 173,228 KB
実行使用メモリ 26,216 KB
最終ジャッジ日時 2024-07-02 11:03:54
合計ジャッジ時間 2,658 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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 * 16
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 #
プレゼンテーションモードにする

// includes
#include <bits/stdc++.h>
// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];
#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size
    ())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = 
    itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; 
    auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto 
    titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << 
    *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << 
    itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); 
    ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef complex<double> cd;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// solve
ll dp[1210][1210];
ll mdp[2][1210];
int at[2][1210][1210];
int curr[2][1210];
int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); INT(m); string s, t; cin >> s >> t;
  rep(i, m)dp[0][i + 1] = 9 + 2 * i;
  rep(i, n)dp[i + 1][0] = 9 + 2 * i;
  rep(i, 2)rep(j, 1201)mdp[i][j] = linf;
  mdp[0][0] = 0;
  mdp[1][0] = 0;
  rep(i, m)mdp[0][i + 1] = dp[0][i+1];
  rep(i, n)mdp[1][i + 1] = dp[i+1][0];
  rep(i, n){
    rep(j, m){
      at[0][i+1][j+1] = -1;
      at[1][i+1][j+1] = -1;
      dp[i+1][j+1] = dp[i][j] + (s[i] == t[j] ? 0: 5);
      dp[i+1][j+1] = min(dp[i+1][j+1], mdp[0][j+1] + 9 + 2 * i);
      dp[i+1][j+1] = min(dp[i+1][j+1], mdp[1][i+1] + 9 + 2 * j);
      if(dp[i+1][j+1] == mdp[0][j+1] + 9 + 2 * i)at[0][i+1][j+1] = curr[0][j+1];
      if(dp[i+1][j+1] == mdp[1][i+1] + 9 + 2 * j)at[1][i+1][j+1] = curr[1][i+1];
      if(mdp[0][j+1] >= dp[i+1][j+1] - 2 * (i + 1)){
        mdp[0][j+1] = dp[i+1][j+1] - 2 * (i + 1);
        curr[0][j+1] = i + 1;
      }
      if(mdp[1][i+1] >= dp[i+1][j+1] - 2 * (j + 1)){
        mdp[1][i+1] = dp[i+1][j+1] - 2 * (j + 1);
        curr[1][i+1] = j + 1;
      }
    }
  }
  cout << dp[n][m] << endl;
  string sr = "", tr = "";
  int nc = n, mc = m;
  while(nc > 0 && mc > 0){
    if(dp[nc][mc] == dp[nc-1][mc-1] + (s[nc-1] == t[mc-1] ? 0: 5)){
      nc--; mc--;
      sr += s[nc];
      tr += t[mc];
    }else if(at[0][nc][mc] >= 0){
      int nct = nc;
      rep(k, nct - at[0][nct][mc])tr += "-";
      rep(k, nct - at[0][nct][mc])sr += s[--nc];
    }else{
      int mct = mc;
      rep(k, mct - at[1][nc][mct])sr += "-";
      rep(k, mct - at[1][nc][mct])tr += t[--mc];
    }
  }
  if(nc > 0)rep(i, nc){
      sr += s[nc-i-1];
      tr += "-";
  }
  if(mc > 0)rep(i, mc){
      tr += t[mc-i-1];
      sr += "-";
  }
  reverse(all(sr)); reverse(all(tr));
  cout << sr << endl;
  cout << tr << endl;
  return 0;
}
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