C++ find the second shortest path using Dijkstras algorithm? - c++

Question

I have a Dijkstras algorithm program below. My program reads a text file and finds the shortest path from the start node to the goal node. My question is once this shorest path is discovered, given my code, how can I store this path? Reason being I am required to also find the second shorest path. Thankyou
#include <iostream>
#include <fstream>
#include <vector>
#include <limits.h>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <utility>
using namespace std;
#define INFINITY1 9999
int counter = 0;
int vcount = 0;
int ecount = 0;
int startVertex;
int endVertex;
int nVerticies;
int nEdges;
struct Vertex
{
int label;
int xcord;
int ycord;
};
struct Edge
{
int parent1;
int parent2;
int weight;
};
Vertex *vertArray = NULL;
Edge *edgeArray = NULL;
float distance(int x1, int y1, int x2, int y2)
{
// Calculating distance
return sqrt(pow(x2 - x1, 2) +
pow(y2 - y1, 2) * 1.0);
}
// Dijkstras
void dijkstra(vector<vector<int>> G, int n, int startnode)
{
// Get x and y cords of start node
int startx = vertArray[startnode].xcord;
int starty = vertArray[startnode].ycord;
// Get x and y cords of end node
int endx = vertArray[n].xcord;
int endy = vertArray[n].ycord;
int **cost = new int *[nVerticies];
for (int i = 0; i < nVerticies; ++i)
{
cost[i] = new int[nVerticies];
}
int distance[nVerticies], pred[nVerticies];
int visited[nVerticies], counter, mindistance, nextnode, i, j;
for (i = 0; i < nVerticies; i++)
for (j = 0; j < nVerticies; j++)
// If there is no weight, then set it to infinity
if (G[i][j] == 0)
cost[i][j] = INFINITY1;
else
{
cost[i][j] = G[i][j];
}
for (i = 0; i < nVerticies; i++)
{
distance[i] = cost[startnode][i];
pred[i] = startnode;
visited[i] = 0;
}
distance[startnode] = 0;
visited[startnode] = 1;
counter = 1;
while (counter < nVerticies - 1)
{
mindistance = INFINITY1;
for (i = 0; i < nVerticies; i++)
if (distance[i] < mindistance && !visited[i])
{
mindistance = distance[i];
nextnode = i;
}
visited[nextnode] = 1;
for (i = 0; i < nVerticies; i++)
if (!visited[i])
if (mindistance + cost[nextnode][i] < distance[i])
{
distance[i] = mindistance + cost[nextnode][i];
pred[i] = nextnode;
}
counter++;
}
for(i=n; i<n+1; i++)
if(i!=startnode)
{
cout<<"\nDistance of node "<<i+1<<"="<<distance[i];
cout<<"\nPath="<<i+1;
j=n;
do
{
j=pred[j+1];
cout<<"<-"<<j+1;
}
while(j!=startnode);
}
}
int main()
{
// Declaration for reading file
int a, b, c;
ifstream readFile;
string fileName;
// Ask for file name
cout << "Enter filename: ";
cin >> fileName;
readFile.open(fileName);
// Read last line and set start and goal vertex equal to file input
while (readFile >> startVertex >> endVertex)
{
}
readFile.close();
readFile.open(fileName);
// Read line 1 and set nVerticies and nEdges equal to file input
while (readFile >> nVerticies >> nEdges)
{
counter++;
break;
}
// Create array with size nVerticies
vertArray = new Vertex[nVerticies];
for (int i = 0; i < nVerticies; i++)
{
// Initialise all elements to zero
vertArray[i].label = 0;
vertArray[i].xcord = 0;
vertArray[i].ycord = 0;
}
// Create array with size nEdges
edgeArray = new Edge[nEdges];
for (int i = 0; i < nEdges; i++)
{
// Initialise all elements to zero
edgeArray[i].parent1 = 0;
edgeArray[i].parent2 = 0;
edgeArray[i].weight = 0;
}
vector<vector<int>> G(nVerticies, vector<int>(nVerticies, 0));
while (readFile >> a >> b >> c)
{
// Get vertex from file and add it to array
if (nVerticies >= counter)
{
vertArray[vcount].label = a;
vertArray[vcount].xcord = b;
vertArray[vcount].ycord = c;
vcount++;
counter++;
}
// Get edge from file and add it to array
else
{
G[a - 1][b - 1] = c;
G[b - 1][a - 1] = c;
edgeArray[ecount].parent1 = a;
edgeArray[ecount].parent2 = b;
edgeArray[ecount].weight = c;
ecount++;
counter++;
}
}
readFile.close();
dijkstra(G, endVertex - 1, startVertex - 1);
int n = 11;
int u = 0;
cout << endl;
cout << endl;
cout << endl;
return 0;
}
output from read file
Distance of node 1=46
Path=1<-2
Distance of node 3=65
Path=3<-1<-2
Distance of node 4=47
Path=4<-2
Distance of node 5=36
Path=5<-2
Distance of node 6=53
Path=6<-2
Distance of node 7=135
Path=7<-8<-6<-2
Distance of node 8=99
Path=8<-6<-2
Distance of node 9=95
Path=9<-4<-2
Distance of node 10=82
Path=10<-2
Distance of node 11=116
Path=11<-10<-2
Distance of node 12=76
Path=12<-2
Distance of node 13=85
Path=13<-2
Ive been guided to follow a similar pattern to the following
if you have a path of 1-->2 -->3 -->4
Set 2 to max and try again
Reset 2
Set 3 to max
Reset 3