結果

問題 No.291 黒い文字列
ユーザー torus711
提出日時 2015-10-16 22:55:35
言語 C++11
(gcc 13.3.0)
結果
TLE  
実行時間 -
コード長 3,696 bytes
コンパイル時間 902 ms
コンパイル使用メモリ 101,740 KB
実行使用メモリ 50,432 KB
最終ジャッジ日時 2024-07-21 20:05:29
合計ジャッジ時間 25,438 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 4
other AC * 13 TLE * 7 -- * 6
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 <iostream>
#include <iomanip>
#include <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <cmath>
#include <cassert>
#include <cstdio>
using namespace std; using namespace placeholders;
using LL = long long;
using ULL = unsigned long long;
using VI = vector< int >;
using VVI = vector< vector< int > >;
using VS = vector< string >;
using SS = stringstream;
using PII = pair< int, int >;
using VPII = vector< pair< int, int > >;
template < typename T = int > using VT = vector< T >;
template < typename T = int > using VVT = vector< vector< T > >;
template < typename T = int > using LIM = numeric_limits< T >;
template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i
    ) << v[i]; } return s; }
template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; };
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); };
#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &e : c )
#define ALL( c ) begin( c ), end( c )
#define AALL( a, t ) ( t* )a, ( t* )a + sizeof( a ) / sizeof( t )
#define DRANGE( c, p ) ( c ).begin(), ( c ).begin() + ( p ), ( c ).end()
#define SZ( v ) ( (int)( v ).size() )
#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter
#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )
#define MP make_pair
#define fst first
#define snd second
#define DUMP( x ) cerr << #x << " = " << ( x ) << endl
map< char, char > nexts = {
{ 'K', 'U' },
{ 'U', 'R' },
{ 'R', 'O' },
{ 'O', 'I' }
};
int dp[ 2 ][ 22 ][ 22 ][ 22 ][ 22 ][ 22 ];
int main()
{
cin.tie( 0 );
ios::sync_with_stdio( false );
string S;
cin >> S;
const int L = SZ( S );
dp[0][0][0][0][0][0] = 1;
REP( i, L )
{
const int cur = i % 2;
const int nex = 1 - cur;
fill( AALL( dp[ nex ], int ), 0 );
REP( a, 21 )
{
REP( b, 21 )
{
REP( c, 21 )
{
REP( d, 21 )
{
REP( e, 21 )
{
dp[ nex ][a][b][c][d][e] |= dp[ cur ][a][b][c][d][e];
if ( S[i] == 'K' || S[i] == '?' )
{
dp[ nex ][ a + 1 ][b][c][d][e] |= dp[ cur ][a][b][c][d][e];
}
if ( ( S[i] == 'U' || S[i] == '?' ) && a )
{
dp[ nex ][ a - 1 ][ b + 1 ][c][d][e] |= dp[ cur ][a][b][c][d][e];
}
if ( ( S[i] == 'R' || S[i] == '?' ) && b )
{
dp[ nex ][a][ b - 1 ][ c + 1 ][d][e] |= dp[ cur ][a][b][c][d][e];
}
if ( ( S[i] == 'O' || S[i] == '?' ) && c )
{
dp[ nex ][a][b][ c - 1 ][ d + 1 ][e] |= dp[ cur ][a][b][c][d][e];
}
if ( ( S[i] == 'I' || S[i] == '?' ) && d )
{
dp[ nex ][a][b][c][ d - 1 ][ e + 1 ] |= dp[ cur ][a][b][c][d][e];
}
}
}
}
}
}
}
int res = 0;
REP( a, 21 )
{
REP( b, 21 )
{
REP( c, 21 )
{
REP( d, 21 )
{
REP( e, 21 )
{
if ( dp[ L % 2 ][a][b][c][d][e] )
{
res = max( res, e );
}
}
}
}
}
}
cout << res << endl;
return 0;
}
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