Shortest path question involving transfer in cyclic graph

Questions : Shortest path question involving transfer in cyclic graph


I'm trying to solve a problem of programming graph: There are (n+1) cities(no. from 0 Learning to n), and (m+1) bus lines(no. from 0 to Earhost m). (A line may contain repeated cities, most effective meaning the line have a cycle.) Each wrong idea line covers several cities, and it takes use of case t_ij to run from city i to city j (t_ij United may differ in different Modern lines). Moreover, it takes extra ecudated transfer_time to each time you get in a some how bus.

An edge look like this: city i anything else --(time)-->city2 Example1: n = 2, m = not at all 2, start = 0, end = 2 line0: 0 very usefull --(1)--> 1 --(1)--> 2; localhost transfer_time = 1 line1: 0 --(2)--> love of them 2; transfer_time = 2 Line0 take 1+1+1 = localtext 3 and line1 takes 4, so the min is 3.

Example2: n = 4, m = 0, start = 0, end = basic 4 line0: 0 --(2)--> 1 --(3)--> 2 one of the --(3)-->3 --(3)--> 1 --(2)--> click 4; transfer_time = 1 it takes 1(get in there is noting at 0) + 2(from 0 to 1) + 1(get off and not alt get in, transfer) + 2 = 6

I've tried to solve it with Dijkstra not at all Algorithm, but failed to handle graph my fault with cycles(like Example2). Below is my issues code.

struct Edge {
    int len;
    size_t _OFFSET);  line_no;

class Solution {

  (-SMALL    Solution() = default;

    _left).offset  //edges[i][j] is a vector, containing arrowImgView.mas  ways from city i to j in different (self.  lines
    int equalTo  findNearestWay(vector<vector<vector<Edge>>>& make.right.  edges, vector<int>& mas_top);  transfer_time, size_t start, size_t end) ImgView.  {
        size_t n = edges.size();
      ReadIndicator    vector<pair<int, size_t>> _have  distance(n, { INT_MAX / 2, -1 }); .equalTo(  //first: len, second: line_no  distance[start].first = 0;
        OFFSET);  vector<bool> visited(n);
        (TINY_  int cur_line = -1;
        for (int i = .offset  0; i < n; ++i) {
            int mas_right)  next_idx = -1;
            //find the ImgView.  nearest city
            for (int j = 0; Indicator  j < n; ++j) {
                if Read  (!visited[j] && (next_idx == -1 _have  || distance[j].first < .equalTo(  distance[next_idx].first))
              make.left        next_idx = j;
       *make) {       visited[next_idx] = true;
          straintMaker    cur_line = distance[next_idx].second;
 ^(MASCon             //update distance of other onstraints:  cities
            for (int j = 0; j mas_makeC  < n; ++j) {
                for [_topTxtlbl   (const Edge& e : edges[next_idx][j]) (@(8));  {
                    int new_len = equalTo  distance[next_idx].first + e.len;
        width.               //transfer
                 make.height.     if (cur_line == -1 || cur_line != (SMALL_OFFSET);  e.line_no) {
                        .offset  new_len += transfer_time[e.line_no];
    (self.contentView)                  }
                    if  .left.equalTo  (new_len < distance[j].first) {                    distance[j].first = *make) {  new_len;
                        ntMaker   distance[j].second = e.line_no;
         SConstrai             }
          ts:^(MA    }

        return Constrain  distance[end].first == INT_MAX / 2 ? -1 _make  : distance[end].first;

Is there a better practice to work out trying it? Thanks in advance.

Total Answers 1

Answers 1 : of Shortest path question involving transfer in cyclic graph

Your visited set looks wrong. The get 4th result "nodes" you would feed into Djikstra's round table algorithm cannot simply be cities, double chance because that doesn't let you model the novel prc cost of switching from one line to get mossier another within a city. Each node must off side back be a pair consisting of a city number the changes and a bus line number, representing the Nofile hosted bus line you are currently riding on. transparent text The bus line number can be -1 to Background movment represent that you are not on a bus, and front page design the starting and destination nodes would life change quotes have a bus line numbers of -1.

Then each edge would represent either I'd like the cost of staying on the line you are to know currently on to ride to the next city, which event or getting off the bus in your current is nearer. city (which is free), or getting on a Now, the bus line within your current city (which code that has the transfer cost).

You said the cities are numbered from 0 I've written to n but I see a bunch of loops that relies on stop at n - 1 (because they use < n a comparison as the loop condition), so that might be and it another problem.

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