这可能吗？
我使用邻接矩阵制作有向加权图。使用矩阵可以有两个权重吗？我可以使用邻接列表可视化如何执行此操作，但这需要重写我的所有代码。
这是我这个项目的全部代码。是的，这是为了学校。不，我不是要求你为我做这项工作。我似乎无法将头围绕在两个砝码上。
所以基本上，我在这里尝试做的是给程序两个权重。我将管理如何进行搜索和实际成本计算，我只是无法弄清楚我应该如何存储这样的两个权重。这两个权重是火车离开的时间，也是火车到达时的时间。之后，我将计算成本/路径/等。基于这两个权重。
我可以拥有一个有两个权重的矩阵吗？我怎么做？我不认为3D矩阵是本课的内容，虽然我认为这样可行（我的想法是有一个3D矩阵，其中z只有2深，所以我可以存储另一个整数＆＃34;后面＆ ＃34;第一个）。但是，如果可能的话，我宁可不要。
正如你所看到的，我误认为任务的目的，并认为只是找到两次之间的差异然后存储，因为权重将起作用。但我需要知道节点x的离开与节点y的到达之间的区别。所以我认为我需要将它们传递到矩阵中然后计算我在做djikstra时获得路径成本。
有什么想法吗？
#include <iostream>
#include <iomanip>
#include <fstream>
#define MAX_SIZE 100
#define INFINITY 999999999
using namespace std;
class graph_matrix
{
public:
graph_matrix();
graph_matrix(int, int);
void setEdge(int, int, int);
void djikstra(int, int);
private:
int nodeCount;
int edgeCount;
int **matrix;
};
void graph_matrix::djikstra(int source, int destination)
{
int distances[nodeCount];
int minNode, minDistance;
bool vertexSet[nodeCount];
for(int i=0; i<nodeCount; i++)
{
distances[i] = INFINITY;
vertexSet[i] = false;
}
distances[source] = 0;
for(int i = 0; i<nodeCount-1; i++)
{
minNode = INFINITY;
minDistance = INFINITY;
for(int vert = 0; vert<nodeCount; vert++)
{
if(vertexSet[vert] == false && distances[vert] <= minDistance)
{
minDistance = distances[vert];
minNode = vert;
}
}
vertexSet[minNode] = true;
for(int vert = 0; vert<nodeCount; vert++)
{
if(vertexSet[vert] == false && matrix[minNode][vert])
{
if(distances[minNode] != INFINITY && distances[minNode] + matrix[minNode][vert] < distances[vert])
{
distances[vert] = distances[minNode] + matrix[minNode][vert];
}
}
}
}
if(distances[destination] < 60)
{
cout << distances[destination] << " minutes." << endl;
}
else if(distances[destination] > 60)
{
cout << (distances[destination] - distances[destination]%60)/60 << " hours and " << distances[destination]%60 << " minutes." << endl;
}
}
graph_matrix::graph_matrix()
{
nodeCount = 0;
edgeCount = 0;
matrix = NULL;
}
graph_matrix::graph_matrix(int nodes, int edges)
{
nodeCount = nodes + 1;
edgeCount = edges;
matrix = new int *[nodeCount];
for (int i = 0; i <= nodeCount; i++)
matrix[i] = new int[nodeCount];
for (int i = 0; i < nodeCount; i++)
{
for (int x = 0; x < nodeCount; x++)
matrix[i][x] = 0;
}
}
void graph_matrix::setEdge(int source, int destination, int weight)
{
matrix[source][destination] = weight;
}
int main()
{
ifstream file;
int statNumber, departStat, arriveStat, departTime, arriveTime, travelTime;
string statName[MAX_SIZE];
graph_matrix *gm = NULL;
file.open("stations.dat");
int stations = 0; //nodes
int tracks = 0; //edges
while(file >> statNumber)
{
file >> statName[statNumber];
stations++;
}
file.close();
file.open("trains.dat");
while(file >> departStat)
{
file >> departStat;//I tried just using 'file;' here, but it didn't work. I didn't see any harm in passing it to a variable, though.
file >> departStat;
file >> departStat;
tracks++;
//I'm sure this is the least elegant way of doing this, but this is what my brain came up with
}
file.close();
file.open("trains.dat");
gm = new graph_matrix(stations, tracks);
while(file >> departStat)
{
file >> arriveStat;
file >> departTime;
file >> arriveTime;
travelTime = arriveTime - departTime; //At first, I thought this is what I needed.
gm->setEdge(departStat, arriveStat, travelTime);
}
file.close();
int menuin, numberin, departin, arrivein;
string stationin;
bool done = false;
cout << "---------------------------------------------" << endl;
cout << " TRAIN MATE 3000 " << endl;
cout << "---------------------------------------------" << endl << endl << endl;
while(!done)
{
cout << "Options" << endl;
cout << "(1) - Look up station by number" << endl;
cout << "(2) - Look up station by name" << endl;
cout << "(3) - Find nonstop train" << endl;
cout << "(4) - Is service available" << endl;
cout << "(5) - Shortest ride time" << endl;
cout << "(6) - Quit" << endl << endl;
cout << "?: ";
cin >> menuin;
cout << endl;
switch (menuin)
{
case 1:
{
cout << ">Enter station number: ";
cin >> numberin;
cout << endl << ">>Station " << numberin << ": " << statName[numberin] << endl << endl;
break;
}
case 2:
{
cout << ">Enter station name: ";
cin >> stationin;
for(int i=0; i<=stations; i++)
{
if(stationin.compare(statName[i]) == 0)
{
cout << endl << ">>Station " << i << ": " << statName[i] << endl << endl;
}
}
break;
}
case 3:
{
cout << ">Enter departure station number: ";
cin >> departin;
cout << ">Enter arrival station number: ";
cin >> arrivein;
//do something to find if there's an edge
break;
}
case 4:
{
cout << ">Enter departure station number: ";
cin >> departin;
cout << ">Enter arrival station number: ";
cin >> arrivein;
//do something to find if there's a path
break;
}
case 5:
{
cout << ">Enter departure station number: ";
cin >> departin;
cout << ">Enter arrival station number: ";
cin >> arrivein;
cout << endl << ">>To go from " << statName[departin] << " to "<< statName[arrivein] << ", you will need to ride the train for ";
gm->djikstra(departin, arrivein);
cout << endl;
break;
}
case 6:
{
done = true;
break;
}
}
}
}
抱歉，我的主要很长。不要担心任何这些。我只是询问矩阵本身，但我想我会给出整个背景。