CF 102697104 - Trans Europe Express

The train network can be modeled as an undirected graph. Every city is a vertex, and every train line is an undirected edge because the line can be traveled in either direction.

codeforcescompetitive-programming

CF 102697104 - Trans Europe Express

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Solution

Problem Understanding

The train network can be modeled as an undirected graph. Every city is a vertex, and every train line is an undirected edge because the line can be traveled in either direction. One particular city is given as the starting point, and we need to count how many distinct cities are reachable from it using any number of train lines. The official problem explicitly suggests BFS and requires mapping city names to numeric identifiers before running the graph traversal.

The first input line gives the number of cities n and train lines m. The next line gives the starting city name. Each of the following m lines contains two city names connected by a train line. The output is the size of the connected component containing the starting city.

The statement does not publish explicit upper bounds for n and m, so we should not assume a small graph. The time limit is only 1 second and the memory limit is 256 MB. That strongly favors a linear graph traversal, O(n + m), rather than repeatedly scanning all train lines while exploring cities. An adjacency list stores exactly the relationships that BFS needs, so both construction and traversal remain linear in the size of the input graph.

City names are strings rather than integer IDs. A direct implementation that tries to use city names as array indices is impossible, so a dictionary is needed to map every encountered name to an integer. Another subtle issue is that a city can occur in several train lines. We must count it once, not once per incident edge.

Consider the following disconnected network.

4 2
A
A B
C D

The correct output is 2. Starting at A, we can reach B, but there is no route from either of them to C or D. An implementation that counts every city appearing in the input would incorrectly return 4.

A second edge case is a cycle.

3 3
A
A B
B C
C A

The correct output is 3. BFS may encounter A again while processing C, but it must not enqueue or count A a second time. Without a visited structure, the traversal can repeatedly walk around the cycle.

The starting city can also have no train lines at all.

1 0
A

The correct output is 1. Being reachable from itself requires zero train rides, so the starting city belongs to the reachable set even when it has no edges.

Approaches

A straightforward but inefficient implementation can represent the train lines as a list of pairs. Whenever BFS processes a city, it scans the entire list of m train lines to find which ones touch that city. This is correct because every possible connection is examined, but processing v reachable cities can require scanning all m edges each time. In the worst case, when all n cities are reachable, that means n * m edge checks. For a dense graph with m on the order of n², this becomes O(n³), and even for a sparse graph it wastes work by repeatedly examining unrelated train lines.

The key observation is that the graph is static. Before BFS starts, we can group the train lines by city. For every city u, store a list containing exactly the cities directly connected to u. Then when BFS reaches u, there is no reason to inspect any unrelated edge. We simply iterate through u's adjacency list.

Because every undirected train line is stored twice, once in each endpoint's adjacency list, the total number of stored adjacency entries is 2m. BFS visits each city at most once and processes each stored adjacency entry at most once. The complete traversal is consequently O(n + m).

The dictionary used for city-name compression is what makes this practical. When a city name is first encountered, assign it the next available integer. All subsequent occurrences of that name refer to the same vertex. The graph algorithm can then operate on integer arrays and lists instead of repeatedly comparing strings.

Approach Time Complexity Space Complexity Verdict
Brute Force O(nm) O(n + m) Too slow
Optimal O(n + m) O(n + m) Accepted

Algorithm Walkthrough

  1. Read n, m, and the starting city name. Create a dictionary that maps city names to integer IDs, and create an adjacency list for the graph. Since the statement gives city names directly rather than numeric vertex labels, the dictionary provides the missing numeric representation.
  2. Read each train line. Convert both endpoint names to integer IDs, creating new IDs when a name is seen for the first time. Add each endpoint to the other's adjacency list because the train line is usable in both directions.
  3. Convert the starting city name to its integer ID. Mark it as visited and put it into a queue. Marking it immediately, rather than waiting until it is removed from the queue, prevents the same city from being inserted multiple times when several already-discovered cities point to it.
  4. Repeatedly remove one city from the front of the queue. Inspect every neighbor in its adjacency list. For every neighbor that has not been visited, mark it visited, increment the reachable-city count, and append it to the queue.
  5. Stop when the queue becomes empty. At that point there is no discovered city with an unexplored outgoing connection, so every city reachable from the starting city has been found. Output the number of visited cities.

Why it works can be expressed as a simple invariant: after every BFS operation, every visited city is reachable from the start, and every reachable city that is at distance at most the portion of the graph already explored has been discovered. When a visited city is processed, every direct neighbor is considered. Thus any reachable city eventually enters the queue. Since a city is marked visited before being enqueued, every city is counted exactly once. When the queue becomes empty, no reachable undiscovered neighbor remains, so the count is exactly the size of the starting city's connected component.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

def solve():
    n, m = map(int, input().split())
    start = input().strip()

    city_id = {}
    graph = [[] for _ in range(n)]

    def get_id(name):
        if name not in city_id:
            city_id[name] = len(city_id)
        return city_id[name]

    start_id = get_id(start)

    for _ in range(m):
        a, b = input().split()

        u = get_id(a)
        v = get_id(b)

        graph[u].append(v)
        graph[v].append(u)

    visited = [False] * len(city_id)
    visited[start_id] = True

    queue = deque([start_id])
    answer = 1

    while queue:
        u = queue.popleft()

        for v in graph[u]:
            if not visited[v]:
                visited[v] = True
                answer += 1
                queue.append(v)

    print(answer)

if __name__ == "__main__":
    solve()

The dictionary city_id performs coordinate compression from arbitrary city names to consecutive integers. get_id is deliberately used for both endpoints of every train line, so equal names always receive the same vertex ID.

The graph is constructed as an undirected adjacency list. The two append operations are both necessary. Omitting the reverse edge would accidentally turn the train network into a directed graph.

The queue contains cities whose adjacency lists still need to be processed. visited[start_id] is set before the first enqueue, and every newly discovered city is marked before being enqueued. This ordering is what prevents duplicates in graphs containing cycles or multiple paths between the same cities.

The answer starts at 1 because the starting city itself is reachable. Every subsequent increment corresponds to one previously unseen city, so the final value is exactly the number of reachable cities.

Python integers do not overflow, and the largest data structures are proportional to the number of city names and train-line endpoints. The implementation also avoids list.pop(0), which would make queue operations O(n). collections.deque.popleft() provides constant-time queue removal.

There is one implementation detail worth handling carefully: the statement gives n, but it does not provide a separate list containing all n city names. The names are discovered from the starting city and the endpoints of the train lines. The adjacency list is initially allocated with n slots, while visited is sized after all names have been encountered. This is safe because the input describes the graph through those names.

Worked Examples

For the first sample, the graph contains one connected component containing Düsseldorf, Frankfurt, Berlin, Hamburg, and Prague. Vienna, Salzburg, and Zurich form a separate component.

Queue operation Current city Newly visited Queue after operation Count
Start Düsseldorf Düsseldorf Düsseldorf 1
1 Düsseldorf Frankfurt Frankfurt 2
2 Frankfurt Berlin Berlin 3
3 Berlin Hamburg, Prague Hamburg, Prague 5
4 Hamburg None Prague 5
5 Prague None Empty 5

The traversal never enters the Vienna-Salzburg-Zurich component because there is no edge connecting it to Düsseldorf's component. The final answer is 5.

For the second sample, Syracuse is connected directly to both Utica and Rochester.

Queue operation Current city Newly visited Queue after operation Count
Start Syracuse Syracuse Syracuse 1
1 Syracuse Utica, Rochester Utica, Rochester 3
2 Utica None Rochester 3
3 Rochester None Empty 3

Both neighbors are discovered while Syracuse is processed. Neither has further undiscovered neighbors, so BFS terminates with all three cities counted.

Complexity Analysis

Measure Complexity Explanation
Time O(n + m) Each city is visited once and each undirected train line contributes two adjacency entries that are inspected once.
Space O(n + m) The dictionary, visited array, queue, and adjacency lists together use linear space.

The official limits give 1 second and 256 MB, while the statement does not publish numerical upper bounds for n and m. A linear solution is the appropriate target because it scales directly with the input size. The adjacency representation also avoids allocating work proportional to the product n * m, which is the main weakness of the brute-force approach.

Test Cases

The following tests use a helper that mirrors the submitted algorithm. The first two are the provided samples. The remaining cases exercise an isolated starting city, a cycle with repeated encounters, a graph in which every city is connected, and a larger chain to catch traversal and boundary mistakes.

import sys
import io
from collections import deque

def solve():
    input = sys.stdin.readline

    n, m = map(int, input().split())
    start = input().strip()

    city_id = {}
    graph = [[] for _ in range(n)]

    def get_id(name):
        if name not in city_id:
            city_id[name] = len(city_id)
        return city_id[name]

    start_id = get_id(start)

    for _ in range(m):
        a, b = input().split()
        u = get_id(a)
        v = get_id(b)
        graph[u].append(v)
        graph[v].append(u)

    visited = [False] * len(city_id)
    visited[start_id] = True

    queue = deque([start_id])
    answer = 1

    while queue:
        u = queue.popleft()

        for v in graph[u]:
            if not visited[v]:
                visited[v] = True
                answer += 1
                queue.append(v)

    print(answer)

def run(inp: str) -> str:
    old_stdin = sys.stdin
    old_stdout = sys.stdout

    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()

    try:
        solve()
        return sys.stdout.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout

# Provided sample 1
assert run(
    """8 7
Dusseldorf
Dusseldorf Frankfurt
Frankfurt Berlin
Berlin Hamburg
Berlin Prague
Vienna Salzburg
Salzburg Zurich
Zurich Vienna
"""
) == "5\n", "sample 1"

# Provided sample 2
assert run(
    """3 2
Syracuse
Syracuse Utica
Syracuse Rochester
"""
) == "3\n", "sample 2"

# Single city with no train lines
assert run(
    """1 0
A
"""
) == "1\n", "isolated starting city"

# Cycle, where every city is reachable but already-visited cities occur again
assert run(
    """4 4
A
A B
B C
C D
D A
"""
) == "4\n", "cycle"

# Two disconnected components, start in the smaller one
assert run(
    """6 4
A
A B
B C
D E
E F
"""
) == "3\n", "disconnected components"

# Larger chain, checking that traversal reaches the final vertex
chain = ["20 19", "C0"]
for i in range(19):
    chain.append(f"C{i} C{i + 1}")

assert run("\n".join(chain) + "\n") == "20\n", "long chain"
Test input Expected output What it validates
1 0, starting city A 1 Minimum-size graph and zero edges
Four vertices forming a cycle 4 Duplicate encounters and visited handling
Two disconnected three-city components 3 Counting only the start component
Twenty-city chain 20 Boundary traversal through the final vertex

Edge Cases

An isolated starting city is the smallest meaningful graph. For

1 0
A

the dictionary assigns A an ID, BFS begins with A, and the queue immediately becomes empty after processing it. The answer remains 1. An implementation that increments the answer only when discovering a neighbor would incorrectly print 0.

A disconnected graph tests whether the traversal follows reachability rather than simply counting input names. For

6 4
A
A B
B C
D E
E F

BFS starts at A, discovers B, then discovers C. There is no edge from this component to D, so D, E, and F are never inserted into the queue. The output is 3.

A cycle exposes duplicate-enqueue bugs. With

4 4
A
A B
B C
C D
D A

BFS first discovers B from A. It then discovers C from B, and D from C. When D is processed, it sees A, but A is already marked visited, so it is ignored. The output is 4, and the traversal terminates instead of looping around the cycle.

Multiple routes to the same city behave similarly. In

4 4
A
A B
A C
B D
C D

both B and C eventually inspect D. Whichever one processes D first marks it visited and enqueues it. The second encounter sees the visited flag and does nothing. The final answer is 4, demonstrating why cities must be marked when discovered rather than counted whenever an edge reaches them.

Finally, a chain such as

5 4
A
A B
B C
C D
D E

checks the traversal boundary. BFS visits A, then B, C, D, and finally E. When E is processed its adjacency list contains only an already visited city, so the queue becomes empty and the answer is 5. An off-by-one implementation that counts only newly discovered neighbors would miss the starting city, while one that stops before processing the final queued vertex could miss E.