URAL1041[Nikifor]

恩,可以看一下LRJ黑书讲贪心那部分,在讲矩阵胚时有提到。
CODE:


/*


PROGRAM: $PROGRAM


AUTHOR: Su Jiao


DATE: 2010-3-24


DESCRIPTION:


$DESCRIPTION


*/


#include <iostream>


using std::cin;


using std::cout;


#include <fstream>


using std::ifstream;


using std::ofstream;


#include <sstream>


using std::stringstream;


using std::endl;


#include <vector>


using std::vector;


#include <string>


using std::string;


#include <stack>


using std::stack;


#include <queue>


using std::queue;


#include <set>


using std::set;


#include <map>


using std::map;


using std::pair;


using std::make_pair;


#include <algorithm>


using std::sort;


#include <cassert>


//using std::assert;


 


class Application


{


      typedef long long int Type;


      int M,N;


      vector<vector<Type> > v;


      vector<pair<int,int> > p;


      static Type abs(Type n)


      {


             return n>=0?n:-n;


      }


      static Type gcd(Type a,Type b)


      {


             return b?gcd(b,a%b):a;


      }     


      bool push(vector<vector<Type> >& get,vector<Type> add)


      {


           vector<int> first(N);


           for (int i=0;i<get.size();i++)


               for (int j=0;j<N;j++)


                   if (get[i][j]!=0)


                   {


                      first[i]=j;


                      break;


                   }


           for (int i=0;i<get.size();i++)


           {


               static const Type BIGPRIME=549979;


               Type _gcd=gcd(abs(add[first[i]]),abs(get[i][first[i]]));


               //assert(_gcd!=0);


               if (_gcd)


               {


               Type mulj=get[i][first[i]]/_gcd;


               Type mulget=add[first[i]]/_gcd;


               for (int j=0;j<N;j++)


                   add[j]=(add[j]*mulj-get[i][j]*mulget)%BIGPRIME;


               }


               bool can=false;


               for (int j=0;j<N;j++)


                   if (add[j]!=0) can=true;


               if (!can) return false;


           }


           get.push_back(add);


           return true;


      }


      public:


      Application()


      {


                    cin>>M>>N;


                    v.resize(M,vector<Type>(N));


                    for (int i=0;i<M;i++)


                        for (int j=0;j<N;j++)


                            cin>>v[i][j];


                    p.resize(M);


                    for (int i=0;i<M;i++)


                    {


                        cin>>p[i].first;


                        p[i].second=i;


                    }


      }


      int run()


      {


          sort(p.begin(),p.end());


          int total=0;


          vector<int> answer;


          vector<vector<Type> > get;


          for (int i=0;i<M;i++)


          {


              if (push(get,v[p[i].second]))


              {


                 total+=p[i].first;


                 answer.push_back(p[i].second);


              }


              if (get.size()==N) break;


          }


          if (get.size()==N)


          {


             cout<<total<<endl;


             sort(answer.begin(),answer.end());


             for (int i=0;i<answer.size();i++)


                 cout<<answer[i]+1<<endl;


          }


          else cout<<0<<endl;


          return 0;


      }


};


 


int main()


{


    Application app;


    return app.run();


}


 


 

留下评论

您的电子邮箱地址不会被公开。 必填项已用 * 标注