Sadonkorjuu

input/code.cpp: In function 'int main()':
input/code.cpp:157:18: warning: 'the_distance' may be used uninitialized in this function [-Wmaybe-uninitialized]
  157 |         g.addEdge(starting_vertex, end_vertex, the_distance);
      |         ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
input/code.cpp:157:18: warning: 'end_vertex' may be used uninitialized in this function [-Wmaybe-uninitialized]
input/code.cpp:157:18: warning: 'starting_vertex' may be used uninitialized in this function [-Wmaybe-uninitialized]

// C++ Program to find Dijkstra's shortest path using
// priority_queue in STL
#include <bits/stdc++.h>
using namespace std; 
#define INF 0x3f3f3f3f
#include<list> 
 
// iPair ==> Integer Pair
typedef pair<int, int> iPair;
 
// This class represents a directed graph using
// adjacency list representation
class Graph {
    int V; // No. of vertices
 
    // In a weighted graph, we need to store vertex
    // and weight pair for every edge
    list<pair<int, int> >* adj;
 
public:
    Graph(int V); // Constructor
 
    //    ction to add an edge to graph
    void addEdge(int u, int v, int w);
 
    // prints shortest path from s
    void shortestPath(int s);
};
 
// Allocates memory for adjacency list
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<iPair>[V];
}
 
void Graph::addEdge(int u, int v, int w)
{
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
}
 
// Prints shortest paths from src to all other vertices
void Graph::shortestPath(int src)
{
    // Create a priority queue to store vertices that
    // are being preprocessed. This is weird syntax in C++.
    // Refer below link for details of this syntax
    // https://www.geeksforgeeks.org/implement-min-heap-using-stl/
    priority_queue<iPair, vector<iPair>, greater<iPair> >
        pq;
 
    // Create a vector for distances and initialize all
    // distances as infinite (INF)
    vector<int> dist(V, INF);
 
    // Insert source itself in priority queue and initialize
    // its distance as 0.
    pq.push(make_pair(0, src));
    dist[src] = 0;
 
    /* Looping till priority queue becomes empty (or all
    distances are not finalized) */
    while (!pq.empty()) {
        // The first vertex in pair is the minimum distance
        // vertex, extract it from priority queue.
        // vertex label is stored in second of pair (it
        // has to be done this way to keep the vertices
        // sorted distance (distance must be first item
        // in pair)
        int u = pq.top().second;
        pq.pop();
 
        // 'i' is used to get all adjacent vertices of a
        // vertex
        list<pair<int, int> >::iterator i;
        for (i = adj[u].begin(); i != adj[u].end(); ++i) {
            // Get vertex label and weight of current
            // adjacent of u.
            int v = (*i).first;
            int weight = (*i).second;
 
            // If there is shorted path to v through u.
            if (dist[v] > dist[u] + weight) {
                // Updating distance of v
                dist[v] = dist[u] + weight;
                pq.push(make_pair(dist[v], v));
            }
        }
    }
 
    // Print shortest distances stored in dist[]
    //printf("Vertex Distance from Source\n");
    for (int i = 0; i < V; ++i)
        printf("%d \t\t %d\n", i, dist[i]);
}
 
// Driver's code
int main()
{
 
    int amount_of_cities; //gets the number of cities that we will have as input
    int amount_of_paths; // we calculate the amount of path between the cities, which is (amount_of_cities-1)
 
    int city_identity_number;
    list<int> city_identifier = {}; // this is the list in which we store the value of 1 and 0 that identifies, if the city has a port or a farm.
 
    list<int> the_list_for_farms = {};   // this is the list that will keep all of the verteces, with the value 1.
    list<int> the_list_for_ports = {};  // this is the list that will keep all the verteces with the value 0.
 
    list<int> the_ordinal_number_of_vertex = {};
 
    int distance_between_2_cities;  // the rows of input that we will get, basically the distance between 2 points.
 
    int starting_vertex;
    int end_vertex;
    int the_distance;
 
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
    cin >> amount_of_cities; //input to see how many verteces we will have.
 
    amount_of_paths = amount_of_cities - 1;
 
    //int array_of_distances[]
 
    Graph g(amount_of_cities);
 
    for (int k = 0; k < amount_of_cities; k++){
        cin >> city_identity_number;
        city_identifier.push_back(city_identity_number);
 
        if (city_identity_number == 1){
            the_list_for_farms.push_back(k);
        }
        if (city_identity_number == 0){
            the_list_for_ports.push_back(k);
        }
    }
 
    for (int i = 0; i < amount_of_paths; i++){
        for (int j = 0; j < 3; j++){
            cin >> distance_between_2_cities;
            if (i == 0){
                starting_vertex = distance_between_2_cities;
                //the_list_for_farms.insert(distance_between_2_cities)
            }
            if (i == 1){
                end_vertex = distance_between_2_cities;
                //the_list_for_ports.insert(distance_between_2_cities)
            }
            if (i == 2){
                the_distance = distance_between_2_cities;
            }
        
        g.addEdge(starting_vertex, end_vertex, the_distance);
            
            
        }
    }
    // create the graph given in above figure
    //int V = 9;
    //Graph g(V);
 
    // making above shown graph
    /*g.addEdge(0, 1, 4);
    g.addEdge(0, 7, 8);
    g.addEdge(1, 2, 8);
    g.addEdge(1, 7, 11);
    g.addEdge(2, 3, 7);
    g.addEdge(2, 8, 2);
    g.addEdge(2, 5, 4);
    g.addEdge(3, 4, 9);
    g.addEdge(3, 5, 14);
    g.addEdge(4, 5, 10);
    g.addEdge(5, 6, 2);
    g.addEdge(6, 7, 1);
    g.addEdge(6, 8, 6);
    g.addEdge(7, 8, 7);
    */
 
    // Function call
    g.shortestPath(0);
 
    return 0;
}

Test 1

Group: 1, 2

Verdict: WRONG ANSWER

Test 2

Group: 1, 2

Verdict: WRONG ANSWER

Test 3

Group: 1, 2

Verdict: RUNTIME ERROR

Test 4

Group: 1, 2

Verdict: RUNTIME ERROR

Test 5

Group: 1, 2

Verdict: RUNTIME ERROR

Test 6

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 7

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 8

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 9

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 10

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 11

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 12

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 13

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 14

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 15

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 16

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 17

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 18

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 19

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 20

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 21

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 22

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 23

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 24

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 25

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 26

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 27

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 28

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 29

Group: 2

Verdict: TIME LIMIT EXCEEDED

Test 30

Group: 1, 2

Verdict: TIME LIMIT EXCEEDED

Test 31

Group: 2

Verdict: TIME LIMIT EXCEEDED