我已经实现了一个只使用整数类型的邻接矩阵图。(考虑到我在这里提到的所有内容的C ++)
我实现了另一个图形,它将使用我的旧实现接收顶点中的城市和边缘中的距离。我想知道这是不是一个好主意,或者我应该使用不同的实现来解决链接列表这个问题。这里的想法是从txt文件中读取所有城市及其相互之间的距离,将其添加到图表中,然后向用户显示菜单,以便他/她可以查询从城市A到B的距离并获取列表他们到达目标之前需要旅行的所有城市。
我计划阅读城市,给它一个数字代码,然后将其添加到图表中,而不是添加字符串" city" (需要将图形从整数转换为字符串)
您如何看待,有什么想法/建议吗?
答案 0 :(得分:0)
您可以使用任一地图。像这样
map<pair<string,string>,int> city ;
(请参阅实现2) 或者,您可以使用vector跟踪城市名称及其索引(实现1)
实施1
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
class Graph
{
int V; // start vertice
map<pair<string,string>,int> city ;
vector<string> city1;
vector<string> city2;
public:
Graph(int start_V)
{
V = start_V;
}
void addEdge(string start,string end,int wt);
void display();
};
void Graph::addEdge(string start,string end,int wt)
{
city[make_pair(start,end)] = wt;
}
void Graph::display()
{
int flag=0;
for(auto it:city)
{
city1.push_back(it.first.first);
city2.push_back(it.first.second);
}
sort(city1.begin(), city1.end());
auto last = unique(city1.begin(), city1.end());
city1.erase(last, city1.end());
sort(city2.begin(), city2.end());
auto last2 = unique(city2.begin(), city2.end());
city2.erase(last2, city2.end());
for(auto col:city2)
{
cout<<"\t"<<col;
}
cout<<"\n";
for(auto row:city1)
{
int flag =0; //for printing row for once
for(auto col:city2)
{
if (!flag)
cout<<row;
cout<<"\t"<<city[make_pair(row,col)];
flag = 1;
}
cout<<"\n";
}
}
int main()
{
Graph g(2);
g.addEdge("A","B",1);
g.addEdge("C","A",5);
g.addEdge("D","E",7);
g.addEdge("E","A",5);
g.addEdge("D","B",7);
g.addEdge("D","L",7);
g.addEdge("W","L",7);
g.display();
return 0;
}
实施2
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
class Graph {
int V; // start vertice
map<pair<string,string>,int> city ;
vector<string> city1;
vector<string> city2;
public:
Graph(int start_V)
{
V = start_V;
}
void addEdge(string start,string end,int wt);
void display(); }; void Graph::addEdge(string start,string end,int wt) { city[make_pair(start,end)] = wt; } void Graph::display() {
int flag=0;
for(auto it:city)
{
city1.push_back(it.first.first);
city2.push_back(it.first.second);
}
sort(city1.begin(), city1.end());
auto last = unique(city1.begin(), city1.end());
city1.erase(last, city1.end());
sort(city2.begin(), city2.end());
auto last2 = unique(city2.begin(), city2.end());
city2.erase(last2, city2.end());
for(auto col:city2)
{
cout<<"\t"<<col;
}
cout<<"\n";
for(auto row:city1)
{
int flag =0; //for printing row for once
for(auto col:city2)
{
if (!flag)
cout<<row;
cout<<"\t"<<city[make_pair(row,col)];
flag = 1;
}
cout<<"\n";
}
} int main() {
Graph g(2);
g.addEdge("A","B",1);
g.addEdge("C","A",5);
g.addEdge("D","E",7);
g.addEdge("E","A",5);
g.addEdge("D","B",7);
g.addEdge("D","L",7);
g.addEdge("W","L",7);
g.display();
return 0; }