Prim’s Algorithm
Time complexity: O(ElogV)
Space complexity: O(V+E)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 |
// Author: Huahua #include <iostream> #include <queue> #include <vector> using namespace std; int main(int argc, char** argv) { const int n = 4; vector<vector<int>> edges{{0,1,1},{0,3,3},{0,2,6},{2,3,2},{1,2,4},{1,3,5}}; vector<vector<pair<int, int>>> g(n); for (const auto& e : edges) { g[e[0]].emplace_back(e[1], e[2]); g[e[1]].emplace_back(e[0], e[2]); } priority_queue<pair<int, int>> q; // (-w, v) vector<int> seen(n); q.emplace(0, 0); // (-w, v) int cost = 0; for (int i = 0; i < n; ++i) { while (true) { const int w = -q.top().first; const int v = q.top().second; q.pop(); if (seen[v]++) continue; cost += w; for (const auto& p : g[v]) { if (seen[p.first]) continue; q.emplace(-p.second, p.first); } break; } } cout << cost << endl; return 0; } |
Python
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 |
# Author: Huahua from collections import defaultdict from heapq import * n = 4 edges = [[0,1,1],[0,3,3],[0,2,6],[2,3,2],[1,2,4],[1,3,5]] g = defaultdict(list) for e in edges: g[e[0]].append((e[1], e[2])) g[e[1]].append((e[0], e[2])) q = [] cost = 0 seen = set() heappush(q, (0, 0)) for _ in range(n): while True: w, u = heappop(q) if u in seen: continue cost += w seen.add(u) for v, w in g[u]: if v in seen: continue heappush(q, (w, v)) break print(cost) |
Kruskal’s Algorithm
Time complexity: O(ElogV)
Space complexity: O(V+E)
C++
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 |
// Author: Huahua #include <iostream> #include <queue> #include <vector> #include <functional> #include <numeric> using namespace std; int main(int argc, char** argv) { const int n = 4; vector<vector<int>> edges{{0,1,1},{0,3,3},{0,2,6},{2,3,2},{1,2,4},{1,3,5}}; vector<vector<int>> q; // (w, u, v) for (const auto& e : edges) q.push_back({e[2], e[0], e[1]}); sort(begin(q), end(q)); vector<int> p(n); iota(begin(p), end(p), 0); function<int(int)> find = [&](int x) { return x == p[x] ? x : p[x] = find(p[p[x]]); }; int cost = 0; for (const auto& t : q) { int w = t[0], u = t[1], v = t[2]; int ru = find(u), rv = find(v); if (ru == rv) continue; p[ru] = rv; // merge (u, v) cost += w; } cout << cost << endl; return 0; } |
Python
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 |
from collections import defaultdict from heapq import * n = 4 edges = [[0,1,1],[0,3,3],[0,2,6],[2,3,2],[1,2,4],[1,3,5]] p = list(range(n)) def find(x): if x != p[x]: p[x] = find(p[p[x]]) return p[x] cost = 0 for u, v, w in sorted(edges, key=lambda x: x[2]): ru, rv = find(u), find(v) if ru == rv: continue p[ru] = rv cost += w print(cost) |
请尊重作者的劳动成果，转载请注明出处！花花保留对文章／视频的所有权利。
如果您喜欢这篇文章／视频，欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Paypal
Venmo
huahualeetcode
huahualeetcode
微信打赏
Be First to Comment