回答編集履歴
2
コード追加
test
CHANGED
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@@ -27,3 +27,223 @@
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１つ以上のグループがレシピと合うやつだけ
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を出力すればよい．
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---
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[追記]
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実装してみたけど，「これはひどい」と言いたくなるコードになった…
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（特に，材料のグループ分け全パターンを探すところがひどい有様）
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```C++
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//EnumGroupingPtn()が使う作業関数
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void EnumPtnWork(
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const std::vector<unsigned int> &Mat,
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std::vector< std::vector<unsigned int> > &Result,
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int idx,
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const std::vector< unsigned int > &Curr
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)
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{
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if( idx == Mat.size() )
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{ Result.push_back( Curr ); return; }
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for( int i=0; i<Curr.size(); ++i )
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{
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std::vector<unsigned int> C = Curr;
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C[i] |= Mat[idx];
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EnumPtnWork( Mat, Result, idx+1, C );
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}
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std::vector<unsigned int> C = Curr;
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C.push_back( Mat[idx] );
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EnumPtnWork( Mat, Result, idx+1, C );
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}
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//材料のグループ分けパターンを全パターン列挙した結果を作る
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std::vector< std::vector<unsigned int> > EnumGroupingPtn( const std::vector<unsigned int> &Materials )
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{
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std::vector< std::vector<unsigned int> > Result;
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EnumPtnWork( Materials, Result, 0, std::vector<unsigned int>() );
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return Result;
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}
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//main
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int main(void)
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{
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//材料の種類要定数：bitパターンで表す
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const unsigned int Stone = 0x01;
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const unsigned int Wood = 0x01 << 1;
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const unsigned int Iron = 0x02 << 2;
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//レシピ
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// key = 作るのに必要な材料の組み合わせ
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// value = <成果物の種類を表す定数(bitパターン), 成果物名称>
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const std::map< unsigned int, std::pair<unsigned int, std::string> > Recipies
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{
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{ Wood, { 0x1, "椅子" } },
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{ Wood|Iron|Stone, { 0x1<<1, "家" } },
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{ Iron, { 0x1<<2, "剣" } },
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{ Wood|Iron, { 0x1<<3, "ノコギリ" } }
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};
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//与えられた材料
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const std::vector<unsigned int> Materials{ Stone, Wood, Iron };
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//結果用データ
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// vector of < 成果物の組み合わせを示す値, 表示用文字列 >
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std::vector< std::pair<unsigned int,std::string> > Result;
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{//メインの処理：材料のグループ分け全パターンについて調査する
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for( const auto &MatGroups : EnumGroupingPtn( Materials ) )
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{
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//このグループ分けで作れる成果物群を調査
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std::string Str_for_Disp; //※これは単なる結果表示用文字列
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unsigned int CraftPtn = 0;
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for( unsigned int MG : MatGroups )
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{
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auto iRecipie = Recipies.find( MG );
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if( iRecipie != Recipies.end() )
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{
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if( CraftPtn>0 )Str_for_Disp += ", ";
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Str_for_Disp += iRecipie->second.second;
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CraftPtn |= iRecipie->second.first;
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}
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}
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if( CraftPtn==0 )continue; //何も作れない場合
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//何か作れる場合
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bool Rejected = false;
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for( auto i=Result.begin(); i!=Result.end(); /*NOP*/ )
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{
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unsigned int And = ( i->first & CraftPtn );
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if( And == CraftPtn ) //既存よりも作れる物が少ない→このパターンを棄却
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{ Rejected=true; break; }
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else if( And == i->first ) //既存よりもたくさん作れる→既存を棄却
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{ i = Result.erase( i ); }
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else
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{ ++i; }
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}
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if( !Rejected ) //棄却されなかったならResultに追加
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{ Result.emplace_back( CraftPtn, Str_for_Disp ); }
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}
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}
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//結果表示
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for( const auto &r : Result )
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{ std::cout << r.second << std::endl; }
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return 0;
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}
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```
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1
修正
test
CHANGED
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@@ -20,4 +20,10 @@
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この中で，
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-
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~~全てのグループに見合うレシピがあるやつだけ~~
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27
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１つ以上のグループがレシピと合うやつだけ
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28
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29
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を出力すればよい．
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