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| 1 | +// Rozwiązanie zadania 'Osiedla' z II etapu XXV OI. |
| 2 | +// Autor rozwiązania: Paweł Putra |
| 3 | +// Złożoność czasowa: O(n) |
| 4 | +// Złożoność pamięciowa: O(n) |
| 5 | +// Punkty: 100 (upsolve) |
| 6 | + |
| 7 | +#include <bits/stdc++.h> |
| 8 | + |
| 9 | +#include <vector> |
| 10 | +#include <iostream> |
| 11 | +#include <algorithm> |
| 12 | +#include <string> |
| 13 | +using namespace std; |
| 14 | +constexpr int MAXN = 1'000'005; |
| 15 | + |
| 16 | +int spojne; |
| 17 | +string kierunki; |
| 18 | +vector<pair<int,int>> g[MAXN]; |
| 19 | + |
| 20 | +int vis[MAXN]; |
| 21 | +int d[MAXN]; // d[v] = odległość v od korzenia (głębokość). |
| 22 | +int low[MAXN]; // low[v] = najniższa głębokość na jaką idzie krawędź powrotna z poddrzewa. |
| 23 | + |
| 24 | +void dfs(int v, int p /* numer krawędzi którą przyszedłem */) { |
| 25 | + low[v] = d[v]; |
| 26 | + vis[v] = 1; |
| 27 | + for (auto [w, i] : g[v]) { |
| 28 | + if (!vis[w]) { |
| 29 | + d[w] = d[v] + 1; |
| 30 | + kierunki[abs(i) - 1] = (i < 0 ? '>' : '<'); |
| 31 | + dfs(w, abs(i)); |
| 32 | + |
| 33 | + low[v] = min(low[w], low[v]); |
| 34 | + } |
| 35 | + else if (abs(i) != p) { |
| 36 | + kierunki[abs(i) - 1] = (i < 0 ? '<' : '>'); |
| 37 | + low[v] = min(low[v], d[w]); |
| 38 | + } |
| 39 | + } |
| 40 | +} |
| 41 | + |
| 42 | + |
| 43 | +int32_t main() { |
| 44 | + ios_base::sync_with_stdio(0); |
| 45 | + int n, m; |
| 46 | + cin >> n >> m; |
| 47 | + |
| 48 | + kierunki = string(m, '<'); |
| 49 | + for (int i = 1; i <= m; i++) { |
| 50 | + int a, b; |
| 51 | + cin >> a >> b; |
| 52 | + g[a].emplace_back(b, i); |
| 53 | + g[b].emplace_back(a, -i); |
| 54 | + } |
| 55 | + |
| 56 | + for (int i = 1; i <= n; i++) { |
| 57 | + if (!vis[i]) |
| 58 | + dfs(i, 0); |
| 59 | + |
| 60 | + if (low[i] == d[i]) spojne++; |
| 61 | + } |
| 62 | + |
| 63 | + cout << spojne << "\n"; |
| 64 | + cout << kierunki << "\n"; |
| 65 | +} |
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