1. Abridged Problem Statement
Given an undirected graph of \(N\) junctions and \(M\) roads. Each junction \(i\) has a traffic light that alternates between blue (duration \(t_{i,B}\)) and purple (duration \(t_{i,P}\)), starting initially in color \(C_i\) with \(r_i\) time units remaining in that color. A vehicle may depart along an edge \((u,v)\) at time \(T\) only if both junction lights at \(u\) and \(v\) share the same color at the instant of departure. Vehicles may wait arbitrarily at any junction. Travel time along edge \((u,v)\) is a fixed integer \(w_{uv}\). Find the minimum arrival time and one corresponding path from source \(S\) to destination \(D\); if none exists, output 0.

2. Detailed Editorial

Overview
We must find a time-dependent shortest-travel path. Standard Dijkstra cannot be applied directly because edge availability depends on departure time through the lights’ phases. However, if at time \(t\) we are at junction \(u\), for each neighbor \(v\) we can compute the earliest time \(\tau \ge t\) at which \(u\) and \(v\) share the same light color, then depart, arriving at time \(\tau + w_{uv}\). We plug those “time-adjusted” edges into a modified Dijkstra.

Key Tasks
1. Modeling Light Phases
   Each junction has an initial phase: color \(C\in\{\text{B,P}\}\), time remaining \(r\), then it alternates indefinitely:
   - If initial \(C=\text{B}\), it stays blue for another \(r\), then purple for \(t_P\), then blue for \(t_B\), purple for \(t_P\), …
   - If initial \(C=\text{P}\), similarly with roles swapped.

2. Querying Color at Time \(t\)
   To decide if \(u\) and \(v\) are both blue (or both purple) at time \(t\), we write a function `get_color(u,t)` returning \(0\) for purple, \(1\) for blue. Internally, we handle the “initial remaining segment” separately, then compute \((t - \text{phaseStart}) \bmod (t_B + t_P)\) to see where in the cycle \(t\) lies.

3. Next-Switch Times
   We need, when colors differ at \(t\), to find the next time either \(u\) or \(v\) switches. We write `time_to_next_color(u,t)` that returns how many time units from \(t\) until junction \(u\) changes color next. We compute that by inspecting the current phase segment and remaining time in it.

4. Synchronizing Two Lights
   Starting at \(t\), we test if `get_color(u,t)==get_color(v,t)`. If yes, \(\tau=t\). If not, we compute \(\Delta_u=time\_to\_next\_color(u,t)\) and \(\Delta_v=time\_to\_next\_color(v,t)\), advance \(t\) by \(\min(\Delta_u,\Delta_v)\), and retry. In the worst case it takes at most a few switches (we cap at 3 attempts) to either find a match or conclude that these two never align in time.

5. Dijkstra with Departure Wait
   - Distances `dist[i]` = earliest known arrival time at \(i\).
   - PQ of \((\text{arrival\_time},\; \text{node})\).
   - Relax an edge \(u\to v\) at current time \(t=dist[u]\) by computing \(\tau=\) first-sync-time\((u,v,t)\). If \(\tau\ge 0\), new arrival time is \(\tau + w_{uv}\). Standard Dijkstra updates if smaller.

6. Path Reconstruction
   Track `parent[v]` = predecessor on the best path. After Dijkstra, walk from destination back to source, then reverse.

Complexities
- Each edge relaxation may advance time by up to 3 phase-change steps, constant work.
- Standard Dijkstra is \(O((N+M)\log N)\). With \(N\le300\), \(M\le14\,000\), this easily fits.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m, source, destination;
vector<tuple<bool, int, int, int>> junctions;
vector<vector<pair<int, int>>> adj;

void read() {
    cin >> source >> destination;
    source--, destination--;
    cin >> n >> m;

    junctions.resize(n);
    for(int i = 0; i < n; ++i) {
        char ci;
        int ri, tb, tp;
        cin >> ci >> ri >> tb >> tp;
        junctions[i] = make_tuple(ci == 'B', ri, tb, tp);
    }

    adj.assign(n, {});
    for(int i = 0; i < m; ++i) {
        int from, to, length;
        cin >> from >> to >> length;
        from--, to--;
        adj[from].push_back({to, length});
        adj[to].push_back({from, length});
    }
}

void solve() {
    // Dijkstra on (junction, time), where a road can only be traversed while
    // both endpoints show the same colour.
    //
    // Each junction is described by (initial colour, remaining time r of that
    // colour, blue duration tB, purple duration tP). get_color returns the
    // colour at absolute time t and time_to_next_color the wait until it next
    // flips; first_time_same_color advances t (waiting at the current
    // junction) until both endpoints agree, returning the earliest such moment
    // or -1 if they never align within a few flips.
    //
    // The Dijkstra state is the arrival time at a junction. To relax an edge we
    // wait for a matching colour, then add the road length. parent[] lets us
    // reconstruct one minimum-time path; if the destination is unreachable we
    // print 0.

    function<int(int, int)> get_color = [&](int u, int t) -> int {
        auto [is_blue, r, tb, tp] = junctions[u];
        if(t < r) {
            return is_blue;
        }

        if(is_blue && t < r + tp) {
            return 0;
        } else if(is_blue) {
            r += tp;
        }

        int cycle = tb + tp;
        int tu = (t - r) % cycle;
        return (int)(tu < tb);
    };

    function<int(int, int)> time_to_next_color = [&](int u, int t) {
        auto [is_blue, r, tb, tp] = junctions[u];
        if(t < r) {
            return r - t;
        }

        int cycle = tb + tp;
        if(is_blue && t < r + tp) {
            return r + tp - t;
        } else if(is_blue) {
            r += tp;
        }

        int tu = (t - r) % cycle;
        return tu < tb ? (tb - tu) : (cycle - tu);
    };

    function<int(int, int, int)> first_time_same_color =
        [&](int u, int v, int t) {
            for(int attempt = 0; attempt < 3; attempt++) {
                if(get_color(v, t) == get_color(u, t)) {
                    return t;
                }

                int dtu = time_to_next_color(u, t);
                int dtv = time_to_next_color(v, t);
                t += min(dtu, dtv);
            }

            return -1;
        };

    vector<int> dist(n, INT_MAX);
    vector<int> parent(n, -1);
    priority_queue<
        pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
        pq;

    pq.push({0, source});
    dist[source] = 0;
    while(!pq.empty()) {
        auto [t, u] = pq.top();
        pq.pop();

        for(auto [v, w]: adj[u]) {
            int first_time = first_time_same_color(u, v, t);
            if(first_time == -1) {
                continue;
            }

            int new_time = first_time + w;
            if(new_time < dist[v]) {
                parent[v] = u;
                dist[v] = new_time;
                pq.push({new_time, v});
            }
        }
    }

    if(dist[destination] == INT_MAX) {
        cout << "0\n";
        return;
    }

    vector<int> path;
    int u = destination;
    while(u != -1) {
        path.push_back(u + 1);
        u = parent[u];
    }

    reverse(path.begin(), path.end());
    cout << dist[destination] << '\n';
    cout << path << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution
```python
import sys
import heapq

def read_input():
    data = sys.stdin.read().split()
    it = iter(data)
    s = int(next(it)) - 1
    d = int(next(it)) - 1
    n = int(next(it))
    m = int(next(it))
    # junctions[i] = (is_blue_initial, rem_initial, dur_blue, dur_purple)
    junctions = []
    for _ in range(n):
        C = next(it)
        r = int(next(it))
        tB = int(next(it))
        tP = int(next(it))
        junctions.append((C == 'B', r, tB, tP))
    adj = [[] for _ in range(n)]
    for _ in range(m):
        u = int(next(it)) - 1
        v = int(next(it)) - 1
        w = int(next(it))
        adj[u].append((v,w))
        adj[v].append((u,w))
    return s, d, n, adj, junctions

def get_color(u, t, junctions):
    is_blue0, rem0, tB, tP = junctions[u]
    # if before initial switch
    if t < rem0:
        return 1 if is_blue0 else 0
    t -= rem0
    cycle = tB + tP
    tt = t % cycle
    # if initial was blue, next block is purple
    if is_blue0:
        if tt < tP:      # in initial purple
            return 0
        tt -= tP
        return 1 if tt < tB else 0
    else:
        if tt < tB:      # in initial blue
            return 1
        tt -= tB
        return 0 if tt < tP else 1

def time_to_next_change(u, t, junctions):
    is_blue0, rem0, tB, tP = junctions[u]
    if t < rem0:
        return rem0 - t
    t -= rem0
    cycle = tB + tP
    tt = t % cycle
    if is_blue0:
        # purple block
        if tt < tP:
            return tP - tt
        tt -= tP
        # blue block
        if tt < tB:
            return tB - tt
    else:
        # blue block
        if tt < tB:
            return tB - tt
        tt -= tB
        # purple block
        if tt < tP:
            return tP - tt
    # fallback
    return cycle - tt

def first_sync_time(u, v, t, junctions):
    # at most 3 trials
    for _ in range(3):
        if get_color(u, t, junctions) == get_color(v, t, junctions):
            return t
        # wait for next change at u or v
        du = time_to_next_change(u, t, junctions)
        dv = time_to_next_change(v, t, junctions)
        t += min(du, dv)
    return -1

def dijkstra(s, d, n, adj, junctions):
    INF = 10**18
    dist = [INF]*n
    parent = [-1]*n
    dist[s] = 0
    pq = [(0, s)]
    while pq:
        t, u = heapq.heappop(pq)
        if t > dist[u]:
            continue
        for v, w in adj[u]:
            depart = first_sync_time(u, v, t, junctions)
            if depart < 0:
                continue
            arrive = depart + w
            if arrive < dist[v]:
                dist[v] = arrive
                parent[v] = u
                heapq.heappush(pq, (arrive, v))
    return dist, parent

def main():
    s, d, n, adj, junctions = read_input()
    dist, parent = dijkstra(s, d, n, adj, junctions)
    if dist[d] == 10**18:
        print(0)
        return
    # reconstruct path
    path = []
    cur = d
    while cur != -1:
        path.append(cur+1)  # 1-based
        cur = parent[cur]
    path.reverse()
    print(dist[d])
    print(*path)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
Treat the problem as a time-dependent shortest-path. For each junction you can query its color at time \(t\) and when it next switches. When at node \(u\) at time \(t\), for each neighbor \(v\) compute the earliest \(t'\ge t\) when both lights match (by stepping to the next switch of \(u\) or \(v\) up to a few times). Then relax with cost \(t' + w_{uv}\). Run Dijkstra, track predecessors. If dest unreachable, print 0; otherwise print min-time and the path.
