समस्या
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.
सीमाएँ
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.
उदाहरण
इनपुट:
cities = 3
routes = [
[1, 2, 5],
[1, 3, 1],
[3, 2, 2]
]
अपेक्षित आउटपुट:
expected = [0, 3, 1]
इनपुट:
long long cities = 3;
std::vector<std::vector<long long>> routes = {
{1, 2, 5},
{1, 3, 1},
{3, 2, 2}
};
अपेक्षित आउटपुट:
std::vector<long long> expected = {0, 3, 1};
इनपुट:
long cities = 3;
long[][] routes = new long[][] {
new long[] {1, 2, 5},
new long[] {1, 3, 1},
new long[] {3, 2, 2}
};
अपेक्षित आउटपुट:
long[] expected = new long[] {0, 3, 1};
व्याख्या
City 3 costs 1. Going from 1 through 3 to 2 costs 1 + 2 = 3, cheaper than the direct cost of 5.
इनपुट:
cities = 1
routes = []
अपेक्षित आउटपुट:
expected = [0]
इनपुट:
long long cities = 1;
std::vector<std::vector<long long>> routes = {};
अपेक्षित आउटपुट:
std::vector<long long> expected = {0};
इनपुट:
long cities = 1;
long[][] routes = new long[][] {};
अपेक्षित आउटपुट:
long[] expected = new long[] {0};
व्याख्या
The only city is the starting city. No travel is needed, so its distance is 0.
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
सारे route costs positive हैं, इसलिए हमेशा अभी तक unreached सबसे सस्ते city को पहले expand करो।
संकेत 2
Directed adjacency lists बनाओ और priority queue से बार-बार outgoing routes relax करो।
संकेत 3
Parallel routes और self-loops ठीक हैं; सिर्फ तब update करो जब नया path सस्ता हो।
अपेक्षित जटिलता
समय O((cities + routes) log cities), अतिरिक्त स्थान O(cities + routes)
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।