結果

問題 No.422 文字列変更 (Hard)
ユーザー 0w1
提出日時 2016-12-05 18:13:35
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 248 ms / 3,000 ms
コード長 3,907 bytes
コンパイル時間 1,604 ms
コンパイル使用メモリ 178,912 KB
実行使用メモリ 307,584 KB
最終ジャッジ日時 2024-11-27 22:32:08
合計ジャッジ時間 5,335 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair< int, int > pii;
typedef vector< int > vi;
typedef vector< vi > vvi;
typedef vector< ll > vl;
typedef vector< vl > vvl;
typedef vector< pii > vp;
typedef vector< vp > vvp;
typedef vector< string > vs;
typedef vector< double > vd;
typedef vector< vd > vvd;
template< class T1, class T2 >
int upmin( T1 &x, T2 v ){
  if( x > v ){
    x = v;
    return 1;
  }
  return 0;
}
template< class T1, class T2 >
int upmax( T1 &x, T2 v ){
  if( x < v ){
    x = v;
    return 1;
  }
  return 0;
}
const int INF = 0x3f3f3f3f;
int N, M;
string S;
string T;
void init(){
  cin >> N >> M;
  cin >> S;
  cin >> T;
}
int dp[ 1200 + 1 ][ 1200 + 1 ][ 3 ][ 3 ];
tuple< int, int, int, int > pre[ 1200 + 1 ][ 1200 + 1 ][ 3 ][ 3 ];
int ds[ 1200 + 1 ][ 1200 + 1 ][ 3 ][ 3 ];
void preprocess(){
  memset( dp, INF, sizeof( dp ) );
  dp[ 0 ][ 0 ][ 0 ][ 0 ] = 0;
  for( int i = 0; i <= N; ++i )
    for( int j = 0; j <= M; ++j )
      for( int t1 = 0; t1 < 3; ++t1 )
        for( int t2 = 0; t2 < 3; ++t2 )
          if( dp[ i ][ j ][ t1 ][ t2 ] < INF ){
            if( i < N and j < M and S[ i ] == T[ j ] ){
              if( upmin( dp[ i + 1 ][ j + 1 ][ 0 ][ 0 ], dp[ i ][ j ][ t1 ][ t2 ] ) )
                pre[ i + 1 ][ j + 1 ][ 0 ][ 0 ] = tie( i, j, t1, t2 ),
                ds[ i + 1 ][ j + 1 ][ 0 ][ 0 ] = 0;
            }
            if( i + 1 <= N and j + 1 <= M ){
              if( upmin( dp[ i + 1 ][ j + 1 ][ 0 ][ 0 ], dp[ i ][ j ][ t1 ][ t2 ] + 5 ) )
                pre[ i + 1 ][ j + 1 ][ 0 ][ 0 ] = tie( i, j, t1, t2 ),
                ds[ i + 1 ][ j + 1 ][ 0 ][ 0 ] = 1;
            }
            if( i + 1 <= N ){/*
              if( upmin( dp[ i + 1 ][ j ][ 1 ][ t2 ], dp[ i ][ j ][ t1 ][ t2 ] + ( t1 != 1 ? 9 : 2 ) ) )
                pre[ i + 1 ][ j ][ 1 ][ t2 ] = tie( i, j, t1, t2 ),
                ds[ i + 1 ][ j ][ 1 ][ t2 ] = 2;*/
              if( upmin( dp[ i + 1 ][ j ][ t1 ][ 2 ], dp[ i ][ j ][ t1 ][ t2 ] + ( t2 != 2 ? 9 : 2 ) ) )
                pre[ i + 1 ][ j ][ t1 ][ 2 ] = tie( i, j, t1, t2 ),
                ds[ i + 1 ][ j ][ t1 ][ 2 ] = 3;
            }
            if( j + 1 <= M ){
              if( upmin( dp[ i ][ j + 1 ][ 2 ][ t2 ], dp[ i ][ j ][ t1 ][ t2 ] + ( t1 != 2 ? 9 : 2 ) ) )
                pre[ i ][ j + 1 ][ 2 ][ t2 ] = tie( i, j, t1, t2 ),
                ds[ i ][ j + 1 ][ 2 ][ t2 ] = 4;/*
              if( upmin( dp[ i ][ j + 1 ][ t1 ][ 1 ], dp[ i ][ j ][ t1 ][ t2 ] + ( t2 != 1 ? 9 : 2 ) ) )
                pre[ i ][ j + 1 ][ t1 ][ 1 ] = tie( i, j, t1, t2 ),
                ds[ i ][ j + 1 ][ t1 ][ 1 ] = 5;*/
            }
          }
}
void solve(){
  int min_cost = INF;
  for( int i = 0; i < 3; ++i )
    for( int j = 0; j < 3; ++j )
      upmin( min_cost, dp[ N ][ M ][ i ][ j ] );
  cout << min_cost << endl;
  for( int i = 0; i < 3; ++i )
    for( int j = 0; j < 3; ++j )
      if( min_cost == dp[ N ][ M ][ i ][ j ] ){
        stack< char > s1, s2;
        for( int a = N, b = M, c = i, d = j; not ( a == 0 and b == 0 ); tie( a, b, c, d ) = pre[ a ][ b ][ c ][ d ] ){
          int op = ds[ a ][ b ][ c ][ d ];
          switch( op ){
            case 0: s1.emplace( S[ a - 1 ] ), s2.emplace( T[ b - 1 ] ); break;
            case 1: s1.emplace( S[ a - 1 ] ), s2.emplace( T[ b - 1 ] ); break;
            case 2: s1.emplace( '-' ); break;
            case 3: s1.emplace( S[ a - 1 ] ), s2.emplace( '-' ); break;
            case 4: s1.emplace( '-' ), s2.emplace( T[ b - 1 ] ); break;
            case 5: s2.emplace( '-' ); break;
          }
        }
        string str1, str2;
        for( ; not s1.empty(); s1.pop(), s2.pop() )
          str1 += s1.top(),
          str2 += s2.top();
        cout << str1 << endl;
        cout << str2 << endl;
        exit( 0 ); 
      }
}
signed main(){
  ios::sync_with_stdio( 0 );
  init();
  preprocess();
  solve();
  return 0;
}
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