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1 | 1 | /** |
2 | | - * Author: Chen Xing |
3 | | - * Date: 2009-10-13 |
| 2 | + * Author: Adam Soltan |
| 3 | + * Date: 2026-01-13 |
4 | 4 | * License: CC0 |
5 | | - * Source: N/A |
6 | 5 | * Description: Fast bipartite matching algorithm. Graph $g$ should be a list |
7 | | - * of neighbors of the left partition, and $btoa$ should be a vector full of |
8 | | - * -1's of the same size as the right partition. Returns the size of |
9 | | - * the matching. $btoa[i]$ will be the match for vertex $i$ on the right side, |
| 6 | + * of neighbors of the left partition, and $r$ should be a vector full of |
| 7 | + * $-1$'s of the same size as the right partition. Returns the size of |
| 8 | + * the matching. $r[i]$ will be the match for vertex $i$ on the right side, |
10 | 9 | * or $-1$ if it's not matched. |
11 | | - * Usage: vi btoa(m, -1); hopcroftKarp(g, btoa); |
12 | | - * Time: O(\sqrt{V}E) |
13 | | - * Status: stress-tested by MinimumVertexCover, and tested on oldkattis.adkbipmatch and SPOJ:MATCHING |
| 10 | + * Time: O(E \sqrt{V}) |
| 11 | + * Status: stress-tested by MinimumVertexCover and tested on Library Checker |
14 | 12 | */ |
15 | 13 | #pragma once |
16 | 14 |
|
17 | | -bool dfs(int a, int L, vector<vi>& g, vi& btoa, vi& A, vi& B) { |
18 | | - if (A[a] != L) return 0; |
19 | | - A[a] = -1; |
20 | | - for (int b : g[a]) if (B[b] == L + 1) { |
21 | | - B[b] = 0; |
22 | | - if (btoa[b] == -1 || dfs(btoa[b], L + 1, g, btoa, A, B)) |
23 | | - return btoa[b] = a, 1; |
24 | | - } |
25 | | - return 0; |
26 | | -} |
27 | | - |
28 | | -int hopcroftKarp(vector<vi>& g, vi& btoa) { |
29 | | - int res = 0; |
30 | | - vi A(g.size()), B(btoa.size()), cur, next; |
31 | | - for (;;) { |
32 | | - fill(all(A), 0); |
33 | | - fill(all(B), 0); |
34 | | - /// Find the starting nodes for BFS (i.e. layer 0). |
35 | | - cur.clear(); |
36 | | - for (int a : btoa) if(a != -1) A[a] = -1; |
37 | | - rep(a,0,sz(g)) if(A[a] == 0) cur.push_back(a); |
38 | | - /// Find all layers using bfs. |
39 | | - for (int lay = 1;; lay++) { |
40 | | - bool islast = 0; |
41 | | - next.clear(); |
42 | | - for (int a : cur) for (int b : g[a]) { |
43 | | - if (btoa[b] == -1) { |
44 | | - B[b] = lay; |
45 | | - islast = 1; |
46 | | - } |
47 | | - else if (btoa[b] != a && !B[b]) { |
48 | | - B[b] = lay; |
49 | | - next.push_back(btoa[b]); |
50 | | - } |
51 | | - } |
52 | | - if (islast) break; |
53 | | - if (next.empty()) return res; |
54 | | - for (int a : next) A[a] = lay; |
55 | | - cur.swap(next); |
| 15 | +int hopcroftKarp(vector<vi>& g, vi& r) { |
| 16 | + int n = sz(g), res = 0; |
| 17 | + vi l(n, -1), q(n), d(n); |
| 18 | + auto dfs = [&](auto f, int u) -> bool { |
| 19 | + int t = exchange(d[u], 0) + 1; |
| 20 | + for (int v : g[u]) |
| 21 | + if (r[v] == -1 || (d[r[v]] == t && f(f, r[v]))) |
| 22 | + return l[u] = v, r[v] = u, 1; |
| 23 | + return 0; |
| 24 | + }; |
| 25 | + for (int t = 0, f = 0;; t = f = 0, d.assign(n, 0)) { |
| 26 | + rep(i,0,n) if (l[i] == -1) q[t++] = i, d[i] = 1; |
| 27 | + rep(i,0,t) for (int v : g[q[i]]) { |
| 28 | + if (r[v] == -1) f = 1; |
| 29 | + else if (!d[r[v]]) d[r[v]] = d[q[i]] + 1, q[t++] = r[v]; |
56 | 30 | } |
57 | | - /// Use DFS to scan for augmenting paths. |
58 | | - rep(a,0,sz(g)) |
59 | | - res += dfs(a, 0, g, btoa, A, B); |
| 31 | + if (!f) return res; |
| 32 | + rep(i,0,n) if (l[i] == -1) res += dfs(dfs, i); |
60 | 33 | } |
61 | 34 | } |
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