Problema
Find the minimum total route cost from city 1 to each city. Routes are directed: [u, v, cost] allows travel from u to v, not automatically back. You may use several routes. City 1 costs 0 to reach. Every city is reachable from city 1.
Implement shortest_weighted_routes(cities, routes).
Input
cities: Number of cities, numbered from 1.
routes: Directed routes [from_city, to_city, cost].
Output
Return the minimum costs in city order, starting with city 1.
Vincoli
1 ≤ cities ≤ 100,000. 0…200,000 routes; endpoints are 1…cities. Costs are integers from 1 to 1,000,000,000. Parallel routes and self-loops are allowed. Distances fit signed 64-bit.
Esempi
Input:
cities = 3
routes = [
[1, 2, 5],
[1, 3, 1],
[3, 2, 2]
]
Output atteso:
expected = [0, 3, 1]Input:
long long cities = 3;
std::vector<std::vector<long long>> routes = {
{1, 2, 5},
{1, 3, 1},
{3, 2, 2}
};
Output atteso:
std::vector<long long> expected = {0, 3, 1};Input:
long cities = 3;
long[][] routes = new long[][] {
new long[] {1, 2, 5},
new long[] {1, 3, 1},
new long[] {3, 2, 2}
};
Output atteso:
long[] expected = new long[] {0, 3, 1};Spiegazione
City 3 costs 1. Going from 1 through 3 to 2 costs 1 + 2 = 3, cheaper than the direct cost of 5.
Input:
cities = 1
routes = []
Output atteso:
expected = [0]Input:
long long cities = 1;
std::vector<std::vector<long long>> routes = {};
Output atteso:
std::vector<long long> expected = {0};Input:
long cities = 1;
long[][] routes = new long[][] {};
Output atteso:
long[] expected = new long[] {0};Spiegazione
The only city is the starting city. No travel is needed, so its distance is 0.
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
Tutti i costi dei percorsi sono positivi, quindi espandi sempre per primo la città non ancora raggiunta più economica.
Suggerimento 2
Costruisci liste di adiacenza dirette e usa una priority queue per rilassare ripetutamente i percorsi uscenti.
Suggerimento 3
Percorsi paralleli e self-loop vanno bene; aggiorna solo quando un nuovo cammino è più economico.
Complessità attesa
tempo O((cities + routes) log cities), spazio aggiuntivo O(cities + routes)
Riferimenti
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