Overview

In a city with \(N\) bus stops and \(M\) bus routes, the problem asks for the minimum number of bus rides needed to travel from bus stop \(S\) to bus stop \(T\).

Analysis

Naive Approach and Its Issues

The most intuitive method is to build a graph where all pairs of bus stops belonging to the same bus route are connected by edges, then find the shortest distance using BFS. However, if bus route \(i\) has \(K_i\) bus stops, the number of pairs is \(O(K_i^2)\). Summing over all routes, the total number of edges becomes extremely large, potentially causing TLE (Time Limit Exceeded).

Key Insight: Introducing Virtual Nodes

To reduce the number of edges, we introduce virtual nodes representing each bus route.

For bus route \(i\), we create a virtual node \(R_i\) and add edges as follows:

• Bus stop → Virtual node (cost 0): The operation of “boarding” a bus route from a bus stop (the ride count does not increase yet since you have only boarded)
• Virtual node → Bus stop (cost 1): The operation of “alighting” at a bus stop from a bus route (one ride is completed)

For example, if route \(i\) passes through bus stops \(\{1, 3, 5\}\):

BusStop1 --0--> R_i --1--> BusStop1
BusStop3 --0--> R_i --1--> BusStop3
BusStop5 --0--> R_i --1--> BusStop5

This way, via “BusStop1 → R_i → BusStop5”, you can travel from BusStop1 to BusStop5 at a cost of \(0 + 1 = 1\). This corresponds to “riding route \(i\) once”.

The number of edges per route is \(2K_i\), so the total is kept to \(O(\sum K_i)\).

Algorithm

Since the edge costs are only of two types: 0 or 1, we use 0-1 BFS instead of regular BFS (which assumes all edge weights are equal).

0-1 BFS is a variant of BFS that uses a double-ended queue (deque):

1. Initialize the distance array \(\mathrm{dist}\) to \(\infty\), and set \(\mathrm{dist}[S] = 0\).
2. Push \(S\) into the deque.
3. Pop a node \(u\) from the front of the deque, and for each adjacent node \(v\):
   • If the edge has cost \(0\), add \(v\) to the front of the deque (processed with priority)
   • If the edge has cost \(1\), add \(v\) to the back of the deque
4. \(\mathrm{dist}[T]\) is the answer. If it remains \(\infty\), output \(-1\).

This allows us to find the shortest distance in \(O(V + E)\) without using a priority queue as in Dijkstra’s algorithm.

Complexity

Let \(K = \sum_{i=1}^{M} K_i\):

• Time complexity: \(O(N + M + K)\)
  • Performing 0-1 BFS on a graph with \(N + M\) nodes and \(2K\) edges
• Space complexity: \(O(N + M + K)\)
  • Required for storing the adjacency list and distance array

Implementation Notes

• Node numbering: It is convenient to number bus stops as \(0, 1, \dots, N-1\) and virtual nodes (route nodes) as \(N, N+1, \dots, N+M-1\).

• Correct usage of 0-1 BFS: Edges with cost 0 should be added using appendleft, and edges with cost 1 using append on the deque. If the order is incorrect, the shortest distances will not be computed correctly.

• Early termination: When \(T\) is popped from the deque, its shortest distance is already finalized, so you can immediately output the answer and terminate.

  Source Code

import sys
from collections import deque

def main():
    data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(data[idx]); idx += 1
    M = int(data[idx]); idx += 1
    S = int(data[idx]) - 1; idx += 1
    T = int(data[idx]) - 1; idx += 1

    # Nodes 0..N-1: bus stops, Nodes N..N+M-1: virtual route nodes
    total = N + M
    adj = [[] for _ in range(total)]

    for i in range(M):
        K = int(data[idx]); idx += 1
        route_node = N + i
        for _ in range(K):
            a = int(data[idx]) - 1; idx += 1
            # stop -> route: weight 0 (boarding)
            adj[a].append((route_node, 0))
            # route -> stop: weight 1 (alighting = 1 ride)
            adj[route_node].append((a, 1))

    # 0-1 BFS
    INF = float('inf')
    dist = [INF] * total
    dist[S] = 0
    dq = deque([S])

    while dq:
        u = dq.popleft()
        d = dist[u]
        if u == T:
            print(d)
            return
        for v, w in adj[u]:
            nd = d + w
            if nd < dist[v]:
                dist[v] = nd
                if w == 0:
                    dq.appendleft(v)
                else:
                    dq.append(v)

    print(dist[T] if dist[T] != INF else -1)

main()

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This editorial was generated by claude4.6opus-thinking.