CF 59E - Shortest Path

Rating: 2000
Tags: graphs, shortest paths
Solve time: 1m 18s
Verified: yes

Solution

Problem Understanding

We are asked to navigate a graph of cities connected by bidirectional roads, with a twist: certain sequences of three consecutive cities are forbidden due to superstition. Formally, there are n nodes and m edges, all unweighted, representing cities and roads. We start at city 1 and want to reach city n along a path that never contains any of the k forbidden triples as consecutive nodes. The output must be the minimal number of roads along such a path and one valid path itself. If no path exists, we return -1.

Given the constraints, we see that n can be up to 3000 and m up to 20000. This makes algorithms with O(n^3) complexity borderline, while O(n^2) algorithms are acceptable. The number of forbidden triples k can reach 10^5, so naive checks for every triple on every potential path will be far too slow.

Non-obvious edge cases arise when the forbidden triples prevent every direct path to the target or when the starting city has only one neighbor, which is part of a forbidden triple sequence. For example, if n = 4, m = 3, roads connect 1-2, 2-3, 3-4, and the forbidden triple is 1-2-3, the naive BFS will incorrectly claim a path exists if it ignores the triple. Another tricky scenario is when a path exists in multiple short forms, and the algorithm must pick the minimal one that avoids all forbidden triples.

Approaches

A brute-force solution would attempt a standard BFS over the graph while remembering the entire path traversed so far. Each time we extend a path by a new city, we check if the last three cities form a forbidden triple. This works in principle, but the number of possible paths grows exponentially with n and m, and storing paths explicitly is prohibitively slow. In the worst case, this naive approach has complexity O(m * 2^n), clearly infeasible for n = 3000.

The key observation is that the forbidden triples only restrict sequences of three consecutive nodes. This means the "memory" we need is only the last two cities visited, not the entire path. If we redefine our BFS state to include the current city and the previous city, we can ensure that every newly visited city does not complete a forbidden triple. With this insight, we can perform a BFS in a state space of size n^2 (previous city × current city), which is manageable given the constraints. The forbidden triples can be stored in a dictionary mapping (prev, current) → set(next forbidden cities) to allow O(1) lookups when extending paths.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(m * 2^n)	O(n * 2^n)	Too slow
BFS with state (prev, current)	O(n * m)	O(n^2 + k)	Accepted

Algorithm Walkthrough

Parse the input to construct the adjacency list for the undirected graph. For each city, we store all directly connected neighbors.
Parse the forbidden triples into a dictionary mapping pairs (prev_city, current_city) to a set of next_city values. This allows O(1) checks when extending the path to see if adding a new city would complete a forbidden triple.
Initialize a BFS queue. Each element of the queue is a tuple (prev_city, current_city). We can start with (0, 1), where 0 represents a dummy city before the start. We also maintain a distance dictionary keyed by (prev_city, current_city) for the shortest path length and a parent dictionary for reconstructing the path.
While the queue is not empty, pop a state (prev_city, current_city) and iterate over all neighbors next_city. If next_city is forbidden after (prev_city, current_city), skip it. If (current_city, next_city) has not been visited, update the distance and parent, then enqueue (current_city, next_city).
If a state (prev_city, n) is reached, reconstruct the path by backtracking through the parent dictionary, starting from (prev_city, n). Collect cities in reverse and then reverse the list to produce the correct order.
If BFS completes without reaching city n, output -1. Otherwise, output the distance and the reconstructed path.

The reason this works is that the BFS guarantees the first time we reach a state (prev_city, n) is through the shortest path that respects forbidden triples. By maintaining only the last two cities in the state, we ensure that we correctly apply all constraints without unnecessary memory usage.

Python Solution

import sys
from collections import deque, defaultdict
input = sys.stdin.readline

n, m, k = map(int, input().split())
graph = [[] for _ in range(n + 1)]
for _ in range(m):
    u, v = map(int, input().split())
    graph[u].append(v)
    graph[v].append(u)

forbidden = defaultdict(set)
for _ in range(k):
    a, b, c = map(int, input().split())
    forbidden[(a, b)].add(c)

queue = deque()
queue.append((0, 1))
distance = dict()
distance[(0, 1)] = 0
parent = dict()

found = False
end_state = None

while queue:
    prev, curr = queue.popleft()
    if curr == n:
        found = True
        end_state = (prev, curr)
        break
    for nxt in graph[curr]:
        if nxt in forbidden.get((prev, curr), set()):
            continue
        if (curr, nxt) not in distance:
            distance[(curr, nxt)] = distance[(prev, curr)] + 1
            parent[(curr, nxt)] = (prev, curr)
            queue.append((curr, nxt))

if not found:
    print(-1)
else:
    path = []
    state = end_state
    while state != (0, 1):
        _, curr = state
        path.append(curr)
        state = parent[state]
    path.append(1)
    path.reverse()
    print(len(path) - 1)
    print(' '.join(map(str, path)))

This solution first builds the graph and forbidden map. BFS then explores the graph while keeping track of the last two cities. The BFS guarantees shortest-path discovery, and the parent dictionary allows us to reconstruct the path efficiently. Using (0, 1) as the initial dummy state ensures uniform handling of forbidden triples at the start.

Worked Examples

Sample Input 1:

Queue	Current State (prev, curr)	Neighbors	New States Enqueued	Distance
[(0,1)]	(0,1)	2,3	(1,2), (1,3)	1
[(1,2),(1,3)]	(1,2)	1,3	(2,3)	2
[(1,3),(2,3)]	(1,3)	1,4	(3,4)	2

We reach (3,4) which is city 4. Backtracking gives path 1 → 3 → 4 with distance 2, avoiding forbidden triple (1,4,3).

Custom Input:

BFS visits (0,1) → (1,2) → cannot go to 3 due to forbidden triple. No path exists. Output -1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n*m)	Each BFS state is (prev, curr) and each edge is checked once per state.
Space	O(n^2 + k)	O(n^2) for distance/parent, O(k) for forbidden triples.

Given n ≤ 3000 and m ≤ 20000, n*m ≈ 6×10^7 operations, feasible within 3 seconds. Memory is within 256 MB.

Test Cases

def run(inp: str) -> str:
    import sys, io
    sys.stdin = io.StringIO(inp)
    from collections import deque, defaultdict
    n, m, k = map(int, input().split())
    graph = [[] for _ in range(n + 1)]
    for _ in range(m):
        u, v = map(int, input().split())
        graph[u].append(v)
        graph[v].append(u)
    forbidden = defaultdict(set)
    for _ in range(k):
        a, b, c = map(int, input().split())
        forbidden[(a, b)].add(c)
    queue = deque()
    queue.append((0, 1))
    distance = dict()
    distance[(0, 1)] = 0
    parent = dict()
    found = False
    end_state = None
    while queue:
        prev, curr = queue.popleft()
        if curr == n:
            found = True
            end_state = (prev, curr)
            break
        for nxt in