From ad12ee1093f71827de632376002504e115aff841 Mon Sep 17 00:00:00 2001
From: Simran <121251182+Simran211103@users.noreply.github.com>
Date: Sun, 29 Oct 2023 00:29:53 +0530
Subject: [PATCH] Create dijkstraApplication.cpp

---
 .../dijkstraApplication.cpp                   | 92 +++++++++++++++++++
 1 file changed, 92 insertions(+)
 create mode 100644 Applications/Computational_Algorithm_Applications/dijkstraApplication.cpp

diff --git a/Applications/Computational_Algorithm_Applications/dijkstraApplication.cpp b/Applications/Computational_Algorithm_Applications/dijkstraApplication.cpp
new file mode 100644
index 0000000..b9fbe7b
--- /dev/null
+++ b/Applications/Computational_Algorithm_Applications/dijkstraApplication.cpp
@@ -0,0 +1,92 @@
+/*
+Dijkstra is a very interesting algorithm that can give us the shortest path reach a certain position.
+Here, we will be considering an example which will be indicating the implemention of Dijkstra's algorithm.
+The example is as stated:
+**Cheapest Flights Within K Stops
+We have n cities and m edges which are connected by some number of flights. 
+1. We have flights array which contains element in the form [fromi, toi, price[i]] i.e we have the subarray in which we have the
+information that we have to go from a given city to another city and the price that is costed by moving from a given city to another city.
+2. We also have three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. 
+3. If we don't find such route, then we can return -1 which signify that we don't have any route that can satisfy all our conditions.
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+  public:
+    int CheapestFLight(int n, vector<vector<int>>& flights, int src, int dst, int K)  {
+        /* we are given the flights so to solve it using Dijkstra's algorithm we have to make an adjacency list to do the further
+         procedure.*/
+        vector<pair<int, int>> adj[n];
+        for (auto it : flights)
+        {
+            adj[it[0]].push_back({it[1], it[2]});
+        }
+
+        /* Now we will create a queue which stores the node and their distances from the 
+        source in the form of {stops, {node, dist}} with ‘stops’ indicating 
+        the number of nodes between src and current node.*/
+        queue<pair<int, pair<int, int>>> q;
+        
+        q.push({0, {src, 0}});
+
+        // Here we are storing the updated distance from our calculations in the code given below.
+        vector<int> dist(n, 1e9);
+        dist[src] = 0;
+
+        // We first iterate through the graph using a queue like in Dijkstra with popping out the element with min stops first.
+        while (!q.empty())
+        {
+            auto it = q.front();
+            q.pop();
+            int stops = it.first;
+            int node = it.second.first;
+            int cost = it.second.second;
+
+            // We stop the process when we the min number of steps we are given.
+            if (stops > K)
+                continue;
+            for (auto iter : adj[node])
+            {
+                int adjNode = iter.first;
+                int edW = iter.second;
+
+                // We only update the queue if the new calculated dist is
+                //less than the prev and the stops are also within limits.
+                if (cost + edW < dist[adjNode] && stops <= K)
+                {
+                    dist[adjNode] = cost + edW;
+                    q.push({stops + 1, {adjNode, cost + edW}});
+                }
+            }
+        }
+       // if after our calculations we still have dist[dist] = 1e9 then we return -1 it signify we can't reach that destination with our
+       // given conditions
+        if (dist[dst] == 1e9)
+            return -1;
+        // otherwise we return the dist[dst] that is the min cost we have to pay to reach from one city to another with min k stops
+        return dist[dst];
+    }
+};
+
+
+int main() {
+        int n; cin>>n;
+        int edge; cin>>edge;
+        vector<vector<int>> flights;
+        
+        for(int i=0; i<edge; ++i){
+            vector<int> temp;
+            for(int j=0; j<3; ++j){
+                int x; cin>>x;
+                temp.push_back(x);
+            }
+            flights.push_back(temp);
+        }
+        
+        int src,dst,k;
+        cin>>src>>dst>>k;
+        Solution obj;
+        cout<<obj.CheapestFLight(n,flights,src,dst,k)<<"\n";
+    return 0;
+}