我正在此代码中实现向导类，但在查找向导家谱中的关系时遇到了严重问题。我正在使用回溯方法来查找两个节点之间的所有关系，但内存将在任务中间开始混乱，我不知道该怎么办。在这里你可以看到这个类使用 C++ 的实现。
#include "Wizard.h"
#include <iostream>
#include <string>
#include <vector>
using namespace std;
Wizard::Wizard(string first_name, string surname, string occupation, string organization, string wand)
: first_name(first_name),surname(surname), occupation(occupation), organization(organization), wand(wand)
{
this->married=0;
this->wand="Dragon Hearstring";
this->occupation="Auror";
this->organization="Dumbledore Army";
};
void Wizard::set_first_name(string _name){
this->first_name = _name;
}
void Wizard::set_surname(string _surname){
this->surname = _surname;
}
void Wizard::set_occupation(string _occupation){
this->occupation = _occupation;
}
void Wizard::set_organization(string _organization){
this->organization = _organization;
}
void Wizard::set_wand(string _wand){
this->wand = _wand;
}
string Wizard::get_name(){
string name;
name = first_name + " " + surname;
return name;
}
string Wizard::get_occupation(){
return occupation;
}
string Wizard::get_organization(){
return organization;
}
string Wizard::get_wand(){
return wand;
}
void Wizard::print_parents(){
cerr << "Parents: ";
for (int i=0; i<parents.size(); i++){
cout << parents[i]->get_name();
if ( i!= parents.size() -1){
cout << " & ";
}
}
cout << endl;
}
void Wizard::print_siblings(){
cerr <<"Siblings: ";
for (int i=0; i<siblings.size(); i++){
cout << siblings[i]->get_name();
if ( i < siblings.size() -1){
cout << " & ";
}
}
cout << endl;
}
void Wizard::print_spouse(){
cerr <<"Spouse: ";
if (married==1)
cout << spouse->get_name();
cout << endl;
}
void Wizard::print_children(){
cerr <<"Children: ";
for (int i=0; i<children.size(); i++){
cout << children[i]->get_name();
if ( i < children.size() - 1){
cout << " & ";
}
}
cout << endl;
}
void Wizard::operator*(Wizard& _spouse){
this->married=1;
_spouse.married=1;
cout << this->get_name() << " and " << _spouse.get_name() << endl;
this->spouse = &_spouse;
_spouse.spouse = this;
_spouse.print_spouse();
vector< Wizard* > these_two;
these_two.push_back(this);
these_two.push_back(&_spouse);
for (int i=0; i<this->children.size(); i++)
children[i]->parents = these_two;
for (int i=0; i<_spouse.children.size(); i++)
children[i]->parents = these_two;
}
void Wizard::operator+(Wizard& _child){
if (!search_in_vector(this->children,&_child))
this->children.push_back(&_child);
if (!search_in_vector(this->spouse->children,&_child))
this->spouse->children.push_back(&_child);
vector< Wizard* > these_two;
these_two.push_back(this);
these_two.push_back(this->spouse);
_child.parents=these_two;
_child.surname = this->surname;
for (int i=0; i<this->children.size(); i++){
if(!search_in_vector(_child.siblings,children[i]) && (children[i]!=&_child)){
_child.siblings.push_back(children[i]);
children[i]->siblings.push_back(&_child);
}
}
}
void Wizard::print_relation_with(Wizard& john_doe){
vector<string> path2;
vector< vector< string > > path1;
vector<Wizard*> nodes_in_the_way;
nodes_in_the_way.push_back(this);
if (this->get_name() == john_doe.get_name()){
cout << this->get_name() << " is " << john_doe.get_name() << endl;
return;
}
this- >search_for_relation(*this,john_doe,path1,path2,nodes_in_the_way);
if (path1.size() == 0){
cout << "no relation" << endl;
return;
}
path2=path1[0];
for (int i=1;i<path1.size();i++){
if (path1[i].size() < path2.size())
path2=path1[i];
}
cout << this->get_name() << " is ";
for (int i=0; i<path2.size(); i++)
cout << path2[i] << " of ";
cout << john_doe.get_name() << endl;
}
void Wizard::search_for_relation(Wizard& a,Wizard& b,vector< vector<string> > &path1,vector<string> &path2,vector< Wizard* > &nodes_in_the_way){
cerr << "////////////" << endl;
cerr << "IM INSIDE " << a.get_name() << " , SEARCHING FOR " << b.get_name() << " - " << path2.size() << endl;
cout << "PATH: ";
for (int i=0;i<nodes_in_the_way.size();i++)
cout << nodes_in_the_way[i]->get_name() << " ";
cout << endl;
a.print_spouse();
a.print_children();
a.print_parents();
if (a.get_name() == b.get_name() && path2.size()==0){
return;
}
if (a.get_name() == b.get_name() ){
path1.push_back(path2);
return;
}
if (a.married){
if (!search_in_vector(nodes_in_the_way,a.spouse)){
path2.push_back("spouse");
nodes_in_the_way.push_back(a.spouse);
search_for_relation(* (a.spouse),b,path1,path2,nodes_in_the_way);
path2.pop_back();
nodes_in_the_way.pop_back();
}
}
// parents
for (int i=0;i<a.parents.size();i++){
if (!search_in_vector(nodes_in_the_way,a.parents[i])){
path2.push_back("child");
nodes_in_the_way.push_back(a.parents[i]);
search_for_relation(* (a.parents[i]),b,path1,path2,nodes_in_the_way);
path2.pop_back();
nodes_in_the_way.pop_back();
}
}
// children
for (int i=0;i<a.children.size();i++){
if (!search_in_vector(nodes_in_the_way,a.children[i])){
path2.push_back("parent");
nodes_in_the_way.push_back(a.children[i]);
search_for_relation(* (a.children[i]),b,path1,path2,nodes_in_the_way);
path2.pop_back();
nodes_in_the_way.pop_back();
}
}
}
///////////// NON-METHOD FUNCITONS ///////////////
bool search_in_vector(vector< Wizard* > a,Wizard* b){
for (int i=0; i<a.size(); i++)
if (a[i]->get_name() == b->get_name())
return true;
return false;
}
这就是代码。在这里你可以看到一些 back_tracking 之后的输出：
在这里您可以看到，Harry 的信息在上次递归期间被弄乱了，可执行文件无法打印他的父母。
请让我知道我做错了！谢谢！