【发布时间】:2011-05-17 21:25:49
【问题描述】:
大家好!
请我尝试根据其中一个数据成员对“类”类型的向量进行排序。如下:
头文件:(Undirected_Graph.h)
#ifndef UNDIRECTED_GRAPH_H
#define UNDIRECTED_GRAPH_H
#include <vector>
using std::vector;
#include <climits>
class Edge;
class Node
{
public:
Node(int); //The constructor.
int id; //For the id of the node.
bool visited; //For checking visited nodes.
int distance;
vector <Edge*> adj; //The adjacent nodes.
};
class Edge
{
public:
Edge(Node*, Node*, int); //The constructor.
Node* start_Node; //The start_Node start of the edge.
Node* end_Node; //The end of the edge.
int w; //The weight of the edge.
bool isConnected(Node* node1, Node* node2) //Checks if the nodes are connected.
{
return((node1 == this->start_Node && node2 == this->end_Node) ||
(node1 == this->end_Node && node2 == this->start_Node));
}
};
class Graph
{
public:
Graph(int); //The Constructor.
int max_Nodes; //Maximum Number of allowed Nodes.
vector <Edge*> edges_List; //For storing the edges of the graph.
vector <Node*> nodes_List; //For storing the nodes of the graph.
void insertEdge(int, int, int);
int getNumNodes();
int getNumEdges();
};
#endif
实施文件:(Undirected_Graph.cpp)
#include "Undirected_Graph.h"
Node::Node(int id_Num)
{
id = id_Num;
visited = 0;
distance = INT_MAX;
}
Edge::Edge(Node* a, Node* b, int weight)
{
start_Node = a;
end_Node = b;
w = weight;
}
Graph::Graph(int size)
{
max_Nodes = size;
for (int i = 1; i <= max_Nodes; ++i)
{
Node* temp = new Node(i);
nodes_List.push_back(temp);
}
}
void Graph::insertEdge(int x, int y, int w)
{
Node* a = nodes_List[x-1];
Node* b = nodes_List[y-1];
Edge* edge1 = new Edge(a, b, w);
Edge* edge2 = new Edge(b, a, w);
edges_List.push_back(edge1);
a->adj.push_back(edge1);
b->adj.push_back(edge2);
}
int Graph::getNumNodes()
{
return max_Nodes;
}
int Graph::getNumEdges()
{
return edges_List.size();
}
现在在上面的代码中,在创建了几个节点和边之后,我需要根据它们的权重对该图的边进行排序。我正在研究一种实现 Kruskal 算法的方法,所以我首先根据边缘的权重对边缘进行排序。
sort (myGraph.edges_List[index].begin(), myGraph.edges_List[index].end());
显然不行!因为向量edges_List 的类型是“Edge”。 假设(myGraph 是该类的一个对象)。
我想知道是否有任何好的技术可以做到这一点?
提前感谢您的帮助!非常感谢任何建议或想法!
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