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 {
public:
(-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
make.top 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) {
make.top 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.
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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