sweden – Swedish train operators that allow assembled bikes

Is there an overview of train lines which carry bicycles without having to disassemble them? I traveled by train from Malmö to Gothenburg last year and it was super easy to take the bicycle with me on the train. However, it’s being said that this is not the case for all train lines in Sweden. Is there an overview on which train lines bicycles can be taken along easily?

public transport – Can bikes be carried on buses in Mallorca?

There are two main providers of public bus services on Mallorca

EMT who provide the city bus routes in Palma.

And, as the question poser highlights, TIB which is the authority that provides bus services across the island.

In the city ‘officially’ EMT will only allow buses on board if they are the folding type

http://www.emtpalma.cat/en/faq

However I have seen people on buses with a bike in the wheelchair space. I guess they would be asked to get off if a wheelchair user needed the spot.

Bus services provided by TIB are actually operated by a number of different companies under licence or contract. And as such means that the are a great number of different bus types. Most of which don’t have a dedicated bike rack or luggage space.

On longer routes such as the ones from Palma to the North and east of the island the service is operated by larger Coach / National Express / Greyhound Type vehicles which have a luggage boot/trunk underneath. I have seen bikes being placed in this space, but there is no means of securing the bikes, so there is a danger of them sliding around. The placing of any luggage / bikes in the storage space is at the discretion of the driver.

In short. City / local buses – officially- folding bikes only

Longer distance coaches. All depends on the vehicle and available space. No specific prohibition, but most definitely at your own risk. And no guarantee that a specific or suitable vehicle will be allocated to any particular service.

Mallorca is an amazing island to cycle, and whilst not in your question, it’s worth mentioning that many car rental services do have bike boxes to rent for an additional fee on top of the car rental price.

automobiles – Which countries don’t have a strong “car culture” (noisy cars & bikes)?

I currently live in Wales. Like most of the UK, it has a very high rate of people driving extremely loud cars and motorbikes, at all times of the day. Like a lot of people, the noise annoys me, but the police do nothing.
It’s like this in many other countries, i.e. Germany.

So my question is, which countries don’t have a high incidence / culture of people driving loud cars, whilst also being relatively well populated? (whilst Greenland would probably not have this, it’s also not a very practical country to live in). Thanks.

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c ++ – LeetCode # 1066 – Campus Bikes II

The following code implements the Hungarian algorithm. The algorithm is used to find the perfect match at minimum cost in a two-part graph.

Look for comments on correctness, efficiency, clarity, and idiomatic C ++ uses. It passes all tests with LeetCode.

class Solution
{
public:
    int assignBikes(vector>& workers, vector>& bikes)
    {
        if (workers.size() == 0)
        {
            return 0;
        }
        if (bikes.size() == 0)
        {
            return 0;
        }
        vector> costs;
        costs.resize(workers.size());
        for (int i = 0; i < workers.size(); i++)
        {
            costs(i).resize(bikes.size());
            for (int j = 0; j < bikes.size(); j++)
            {
                int wx = workers(i)(0);
                int wy = workers(i)(1);
                int bx = bikes(j)(0);
                int by = bikes(j)(1);
                int dx = wx - bx;
                int dy = wy - by;
                dx = dx > 0 ? dx : -dx;
                dy = dy > 0 ? dy : -dy;
                costs(i)(j) = dx + dy;
            }
        }
        vector> answer = hungarian(costs);
        int result = 0;
        for (auto edge : answer)
        {
            result += costs(edge.first)(edge.second);
        }
        return result;
    }
private:
    // Implementation of the Hungarian algorithm for finding a minimum cost perfect matching
    // The implementation is based on the description on https://en.wikipedia.org/wiki/Hungarian_algorithm
    vector> hungarian(const vector>& costs)
    {
        // The problem of finding a minimum cost assignment is the same as
        // finding the minimum cost perfect matching between the rows (workers) and the columns (jobs)
        int n = (int)costs.size();
        int m = (int)costs(0).size();

        // At all times, the algorithm maintains an integer named potential for all nodes.
        // The algorithm make sure potential(u) + potential(v) <= cost(u,v)
        // The nodes that represents the row are (0,n) and the nodes that represents the columns are (n, m + n)
        // Potential values are initialized to 0.
        vector potential;
        potential.resize(m + n);
        for (int i = 0; i < m + n; i++)
        {
            potential(i) = 0;
        }

        // We maintain a directed graph, the edges of the graph is tight (i.e. potential(u) + potential(v) = cost(u,v)).
        // The edges from (n, m + n) back to (0, n) has to form a matching. (i.e. these edges does not share source/target)
        vector> direction;
        direction.resize(n);
        for (int i = 0; i < n; i++)
        {
            direction(i).resize(m);
            for (int j = 0; j < m; j++)
            {
                direction(i)(j) = true;
            }
        }

        // If a node is involved in the matching, the matched flag will be true.
        vector matched;
        matched.resize(m + n);
        for (int i = 0; i < m + n; i++)
        {
            matched(i) = false;
        }

        // We will do a BFS to find paths. The parent array store the parent node used to discover a node.
        vector parent;
        parent.resize(m + n);

        while (true)
        {
            queue bfs;
            for (int i = 0; i < m + n; i++)
            {
                // Parent == -1 means it is not enqueued yet, so there is no parent
                parent(i) = -1;
            }

            for (int i = 0; i < n; i++)
            {
                if (!matched(i))
                {
                    bfs.push(i);
                    // Parent == -2 means it is enqueued because it is a not matched source node.
                    // It should NOT be enqueued again
                    parent(i) = -2;
                }
            }

            // This is just a standard BFS, note that the cost and direction array must be read from row to column
            while (bfs.size() > 0)
            {
                int visiting = bfs.front();
                bfs.pop();
                if (visiting < n)
                {
                    for (int neighbor = n; neighbor < m + n; neighbor++)
                    {
                        bool tight = costs(visiting)(neighbor - n) - potential(visiting) - potential(neighbor) == 0;
                        if (direction(visiting)(neighbor - n) && tight)
                        {
                            if (parent(neighbor) == -1)
                            {
                                bfs.push(neighbor);
                                parent(neighbor) = visiting;
                            }
                        }
                    }
                }
                else
                {
                    for (int neighbor = 0; neighbor < n; neighbor++)
                    {
                        bool tight = costs(neighbor)(visiting - n) - potential(visiting) - potential(neighbor) == 0;
                        if (!direction(neighbor)(visiting - n) && tight)
                        {
                            if (parent(neighbor) == -1)
                            {
                                bfs.push(neighbor);
                                parent(neighbor) = visiting;
                            }
                        }
                    }
                }
            }

            bool found = false;
            for (int i = n; i < m + n; i++)
            {
                if (!matched(i) && parent(i) != -1)
                {
                    // If there is a way to reach from a source unmatched node to a target
                    // unmatched node, then we have an augmenting path, using the parent
                    // pointer chain, we can reverse all the paths to increasing the size
                    // of the matching by 1.
                    found = true;
                    int current = i;
                    while (true)
                    {
                        if (parent(current) == -2)
                        {
                            matched(i) = true;
                            matched(current) = true;
                            break;
                        }
                        int small = min(current, parent(current));
                        int large = max(current, parent(current));
                        direction(small)(large - n) = !direction(small)(large - n);
                        current = parent(current);
                    }
                    break;
                }
            }

            if (!found)
            {
                // If we cannot find an augmenting path, we will modify the potential
                // so that we have more nodes reachable from the unmatched sources
                int delta = -1;
                for (int i = 0; i < n; i++)
                {
                    if (parent(i) != -1)
                    {
                        for (int j = n; j < m + n; j++)
                        {
                            if (parent(j) == -1)
                            {
                                // Make sure the delta does not violate the constraint
                                // between each pair of reachable source/unreachable target
                                int margin = costs(i)(j - n) - potential(i) - potential(j);
                                if (delta == -1 || margin < delta)
                                {
                                    delta = margin;
                                }
                            }
                        }
                    }
                }
                if (delta == -1)
                {
                    break;
                }
                for (int i = 0; i < n; i++)
                {
                    if (parent(i) != -1)
                    {
                        // For each reachable source, the potential is increased by delta
                        potential(i) += delta;
                    }
                }
                for (int i = n; i < m + n; i++)
                {
                    if (parent(i) != -1)
                    {
                        // For each reachable target, the potential is decreased by delta
                        potential(i) -= delta;
                    }
                }
                // After the manipulation:
                // The potential is feasible for reachable source, reachable target because the sum didn't change.
                // The potential is feasible for reachable source, unreachable target because the increase is checked earlier.
                // The potential is feasible for unreachable source, reachable target because the sum decrease.
                // The potential is feasible for unreachable source, unreachable target because the sum didn't change.
            }
        }
        vector> answer;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (!direction(i)(j))
                {
                    answer.push_back(make_pair(i, j));
                }
            }
        }
        return answer;
    }
};