我试图代表一个无方向图。我创建了三个结构,如下所示。我将运算符==和运算符
struct Node; /* Forward references to these two types */
struct Edge; /* that the compiler can recognize them */
/* Type: Node
* This type represents an individual node and consists of the data of the
* node and the set of edges from this node */
struct Node{
int nodeNum;
string data;
set<Edge*> edges;
};
/* Type: Edge
* This type represents an individual edge and consists of pointers to the
* endpoints */
struct Edge{
Node *end1;
Node *end2;
// This says that edge from node 1 to node 2 and edge from node 2 to node 1 are considered the same
bool operator==(const Edge &e) const{
return ( (this->end1->nodeNum == e.end1->nodeNum && this->end2->nodeNum == e.end2->nodeNum) ||
(this->end1->nodeNum == e.end2->nodeNum && this->end2->nodeNum == e.end1->nodeNum));
}
// This function is used by set to order elements of edges.
bool operator<(const Edge *e) const{
return (this->end1 < e->end1 && this->end2 < e->end2);
}
};
// This is a struct for graph
struct Graph{
set<Node*> Nodes;
set<Edge*> Edges;
map<int, Node*> nodeMap;
};
问题:如果说,我从节点1到2有一条边,从2到1有一条边,我的struct声明说它们应该被认为是等效的。但是,当我将这两个边缘插入到集合中时,它将两个元素都插入为两个单独的元素(即集合不了解1-2和2-1边缘相等)。我该怎么做才能使集合处理重复项(即仅保留这些边缘之一)。参见例如下面:
int main(){
// Let's make 2 nodes, node 1 and node 2
Node* n1 = new Node;
Node* n2 = new Node;
n1->nodeNum=1;
n2->nodeNum=2;
// Let's make 2 edges 1-2 and 2-1
Edge* e1 = new Edge;
Edge* e2 = new Edge;
e1->end1=n1; e1->end2=n2;
e2->end1=n2; e2->end2=n1;
// Now let's make a graph and put the edges in its internal set
Graph g;
g.Edges.insert(e1);
g.Edges.insert(e2); // the set takes in both e1 and e2. If I print all elements in g.Edges, it will print both 1-2 and 2-1
// How do I tell the set to treat e1 and e2 as equal edges so it took care of duplicates?
return 0;
}
最佳答案
std::set<T*>
将创建一组存储位置,而不是一组T值。
如果要比较指向的对象,则需要提供一个自定义比较器:
struct Ptr_compare {
template<typename T>
constexpr bool operator()( const T* lhs, const T* rhs ) const {
return *lhs < *rhs;
}
};
// This is a struct for graph
struct Graph {
set<Node*, Ptr_compare> Nodes;
set<Edge*, Ptr_compare> Edges;
map<int, Node*> nodeMap;
};
然而:
请注意,我编写的代码可以回答您的问题,但是对于您的用例仍然不正确,只能将其用于没有所有权的指针,这绝对不是您的情况。
这不是我的解决方案本身的问题,而是您要完成的工作的根本问题。需要对去重复的对象调用
delete
。关于c++ - 使用一组结构,并避免一组重复的结构,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/52747379/