Minimum Fuel Cost to Report to the Capital

class Solution {
public:
    long long computeFuel(int cities, vector<vector<int>>& connections, vector<int>& cityDegrees, int& maxSeats) {
        queue<int> processingQueue;
        for (int id = 1; id < cities; id++) {
            if (cityDegrees[id] == 1) {
                processingQueue.push(id);
            }
        }

        vector<int> citizenCount(cities, 1);
        long long totalFuel = 0;

        while (!processingQueue.empty()) {
            int current = processingQueue.front();
            processingQueue.pop();

            totalFuel += ceil((double)citizenCount[current] / maxSeats);
            for (auto& adjacent : connections[current]) {
                cityDegrees[adjacent]--;
                citizenCount[adjacent] += citizenCount[current];
                if (cityDegrees[adjacent] == 1 && adjacent != 0) {
                    processingQueue.push(adjacent);
                }
            }
        }
        return totalFuel;
    }

    long long minFuel(vector<vector<int>>& routes, int maxSeats) {
        int cityCount = routes.size() + 1;
        vector<vector<int>> adjacencyList(cityCount);
        vector<int> cityDegree(cityCount);

        for (auto& route : routes) {
            adjacencyList[route[0]].push_back(route[1]);
            adjacencyList[route[1]].push_back(route[0]);
            cityDegree[route[0]]++;
            cityDegree[route[1]]++;
        }

        return computeFuel(cityCount, adjacencyList, cityDegree, maxSeats);
    }
};

The C++ solution involves finding the minimum fuel cost required for citizens from various cities to report to a capital using available routes, where the amount of people a vehicle can transport is limited by maxSeats. The code defines a Solution class with two primary methods: minFuel and computeFuel.

Implementation Overview:

• minFuel initializes the problem:

  • It calculates the number of cities.
  • It builds an adjacency list and a degree list for each city using the provided routes.
  • It then calls computeFuel passing the constructed lists and the maxSeats parameter.

• computeFuel calculates the minimum fuel cost:

  • It initializes a queue to process cities based on their connections.
  • It processes each city by dequeuing and calculating the fuel cost based on the citizens count and maxSeats.
  • It updates the citizen count and city connections as it progresses through each city.

Details on Data Structures and Approach:

• Uses a vector for the adjacency list (vector<vector<int>>) and citizen counts, and an queue<int> for cities to be processed.
• The city degree is reduced as connections are processed, and a city is added to the process queue when it has only one remaining connection (indicating a direct route).
• The fuel cost is calculated using the ceiling value of citizens in a city divided by maxSeats, ensuring any remainder of citizens also requires transport.
• Each city's calculations contribute cumulatively to the totalFuel which is ultimately returned.

This method effectively models the problem as a leaf-node initiation of breadth-first traversal, cascading the count of people through the graph towards the capital, and calculating required fuel based on groups of people allowed per transport. The solution essentially reduces the problem down to understanding and manipulating basic graph traversal and aggregation operations.

class Solution {
    public long calculateFuel(int cityCount, Map<Integer, List<Integer>> graph, int[] connectionCount, int busCapacity) {
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 1; i < cityCount; i++) {
            if (connectionCount[i] == 1) {
                queue.offer(i);
            }
        }

        int[] peopleCount = new int[cityCount];
        Arrays.fill(peopleCount, 1);
        long totalFuel = 0;

        while (!queue.isEmpty()) {
            int current = queue.poll();
            totalFuel += Math.ceil((double) peopleCount[current] / busCapacity);

            for (int adjacent : graph.get(current)) {
                connectionCount[adjacent]--;
                peopleCount[adjacent] += peopleCount[current];
                if (connectionCount[adjacent] == 1 && adjacent != 0) {
                    queue.offer(adjacent);
                }
            }
        }
        return totalFuel;
    }

    public long findMinFuel(int[][] connections, int busCapacity) {
        int nodeCount = connections.length + 1;
        Map<Integer, List<Integer>> adjacencyList = new HashMap<>();
        int[] degree = new int[nodeCount];

        for (int[] connection : connections) {
            adjacencyList.computeIfAbsent(connection[0], k -> new ArrayList<Integer>()).add(connection[1]);
            adjacencyList.computeIfAbsent(connection[1], k -> new ArrayList<Integer>()).add(connection[0]);
            degree[connection[0]]++;
            degree[connection[1]]++;
        }

        return calculateFuel(nodeCount, adjacencyList, degree, busCapacity);
    }
}

The provided Java code outlines a solution for calculating the minimum fuel cost required to transport people from various cities to a capital city, given constraints on bus capacity and city connectivity. The approach used here involves working with an adjacency list representation of cities connected by roads (edges) and performing calculations to determine effective fuel consumption.

Here's how the solution is properly structured:

• Initialization and Building Adjacency List: In the findMinFuel method, the adjacency list is constructed using the given city connections. Each city pair in the connections array represents a bidirectional road, and for each city, increments the connection count which represents the number of roads connecting to that city.

• BFS (Breadth-first search) Calculation: The calculateFuel method uses a BFS approach, aided by a queue. Begin by enqueuing all cities that are connected by only one road (besides the capital, assumed here to be represented by index 0 with a special condition). These cities are starting points for the fuel calculation iterations since their situation often involves direct transit requirements without intermediate stops.

• Fuel Calculation Process: During the BFS, for each city dequeued:

  • Calculate fuel based on current people count and bus capacity using the ceil function to ensure whole bus trips.
  • Adjust the people count and connection counts of adjacent cities.
  • If an adjacent city's connections reduce to one (indicating it has one direct route remaining potentially leading to the capital), enqueue this city for upcoming calculation.

This procedure continues until no more cities need to be reviewed in the queue. The cumulative fuel used is calculated and returned as the minimum required fuel cost.

Overall, this method efficaciously combines graph traversal techniques with a practical application scenario, allowing it to efficiently determine minimum fuel requirements under given constraints.