逆向图

问题描述

如何反转不转置定向图？

 Graphe *Graphe::grapheInverse( void ){
     Graphe *r = new Graphe (_adjacences.size() );
      for (unsigned i = 0; i < _adjacences.size(); i++){
          for (unsigned j = 0; j < _adjacences[i]->size(); j++){
              //r->addArcs(j,i); this doesn't work
          }
      }
      return r;
  }

_adjacences为vector< vector< int > * > _adjacences;

和addArcs是：

void Graphe::addArcs( int a_sommet1,int a_sommet2 ){
    assert( 0 <= a_sommet1 && a_sommet1 < _adjacences.size() );
    assert( nullptr != _adjacences[a_sommet1] );
    _adjacences[a_sommet1]->push_back( a_sommet2 );
}

示例：

    Graphe g( 5 );
    g.addArcs( 0,1 );
    g.addArcs( 0,4 );
    g.addArcs( 1,0 );
    g.addArcs( 1,4 );
    g.addArcs( 2,0 );
    g.addArcs( 2,1 );
    g.addArcs( 2,3 );
    g.addArcs( 2,4 );
    g.addArcs( 4,3 );
    g.addArcs( 4,1 );

    // inversion du graphe :
    Graphe *r = g.grapheInverse();

我的完整代码如下：

Graphe.hpp

#include <iostream>
#include <vector>

using namespace std;

class Graphe
{
private:
    vector< vector< int > * > _adjacences;
public:
    Graphe( void );
    Graphe( int a_nbSommet );
    virtual ~Graphe( void );

    int nbSommet( void ) const;
    vector< int > * adjacences( int a_sommet );
    void addArcs( int a_sommet1,int a_sommet2 );

    Graphe * grapheInverse( void );

    friend ostream & operator <<( ostream &,Graphe const & );
};

Graphe.cpp

#include "Graphe.hpp"

#include <algorithm>
#include <cassert>
#include <iomanip>
#include <iostream>
#include <vector>

using namespace std;

Graphe::Graphe( void )
{
}


Graphe::Graphe( int a_nbSommet ): _adjacences( a_nbSommet ){
    int i;
    for( i = 0; i < a_nbSommet; ++ i ){
        _adjacences[i] = new vector< int >();
    }
}


Graphe::~Graphe( void )
{
}


void Graphe::addArcs( int a_sommet1,int a_sommet2 ){
    assert( 0 <= a_sommet1 && a_sommet1 < _adjacences.size() );
    assert( nullptr != _adjacences[a_sommet1] );
    _adjacences[a_sommet1]->push_back( a_sommet2 );
}


int Graphe::nbSommet( void ) const{
    return _adjacences.size();
}


vector< int > *Graphe::adjacences( int a_sommet ){
    assert( 0 <= a_sommet && a_sommet < _adjacences.size() );
    return _adjacences[ a_sommet ];
}


Graphe *Graphe::grapheInverse( void ){
    Graphe *r = new Graphe (_adjacences.size() );
    for (unsigned i = 0; i < _adjacences.size(); i++)
        for ( unsigned j = 0; j < _adjacences[i]->size(); j++ )
            r->addArcs ((*_adjacences[i])[j],i); 
    return r;
}


ostream &operator <<( ostream & a_out,Graphe const & a_graphe ){
    int i = 0;
    int j = 0;

    for( i = 0; i < a_graphe._adjacences.size(); ++ i ){
        a_out << i << " : ";
        for( j = 0; j < a_graphe._adjacences[i]->size(); ++ j ){
            if( 0 != j ){
                a_out << ",";
            }
            a_out << ( a_graphe._adjacences[i]->at(j) );
        }
        a_out << endl;
    }

    return a_out;
}

我的主要人

#include "Graphe.hpp"


#include <algorithm>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <vector>

using namespace std;

int main( int argn,char * argv[] ){
    // construction of the graph :
    Graphe g( 5 );
    g.addArcs( 0,1 );

    // inversion of the graph :
    Graphe *r = g.grapheInverse();

    // printing the both lists for verification
    cout << g << endl;
    cout << *r << endl;
}

这给了我

g:
0 : 1,4
1 : 0,4
2 : 0,1,3,4
3 : 
4 : 3,1

r:
0 : 1,2
1 : 0,2,4
2 : 
3 : 2,4
4 : 0,2

解决方法

制作反向列表时，您不想将原始列表中的索引用作源顶点，因此需要取消引用列表。所以您要使用

r->addArcs((*_adjacences[i])[j],i);

有更好的方法来编写该函数（例如基于范围的for循环）。