제출 #934771

#제출 시각아이디문제언어결과실행 시간메모리
934771eysbutnoHard route (IZhO17_road)C++17
19 / 100
2 ms1112 KiB
#include <bits/stdc++.h> using namespace std; using ll = long long; using pii = pair<int, int>; #define all(x) begin(x), end(x) #define sz(x) (int) (x).size() #define f first #define s second template<class T> bool smin(T& a, T b) { return b < a ? a = b, 1 : 0; } template<class T> bool smax(T& a, T b) { return b > a ? a = b, 1 : 0; } int main() { cin.tie(0) -> sync_with_stdio(0); int n; cin >> n; vector adj(n, vector<int>()); for (int i = 1; i < n; i++) { int x, y; cin >> x >> y; adj[--x].push_back(--y); adj[y].push_back(x); } vector<int> d(n), c(n); auto dfs = [&](int u, int p, auto&& dfs) -> void { d[u] = 0, c[u] = 1; for (int v : adj[u]) if (v != p) { dfs(v, u, dfs); if (smax(d[u], d[v] + 1)) { c[u] = c[v]; } else if (d[v] + 1 == d[u]) { c[u] += c[v]; } } }; dfs(0, -1, dfs); ll hard = 0, cnt = 1; auto dfs2 = [&](int u, int p, int pd, int pc, auto&& dfs2) -> void { vector<pii> opt; if (u > 0 || sz(adj[u]) == 1) { opt.push_back({pd, pc}); } for (int v : adj[u]) if (v != p) { opt.push_back({d[v] + 1, c[v]}); } sort(all(opt), greater<>()); if (sz(adj[u]) >= 3) { // can form nonzero path ll cur = opt[0].f * (opt[1].f + opt[2].f), num = 0; int ties = 0; // equal to third element for (auto [k, v] : opt) { if (k == opt[2].f) ties += v; } // case 1: all are distinct if (opt[0].f != opt[1].f && opt[1].f != opt[2].f) { // num = opt[0].s * opt[1].s * ties; num = opt[1].s * ties; } // case 2: all are the same else if (opt[0].f == opt[1].f && opt[1].f == opt[2].f) { // n choose 2, in this case num = ties * (ties - 1) / 2; } // case 3: first two are the same else if (opt[0].f == opt[1].f) { num = (opt[0].s + opt[1].s) * ties; } // case 4: last two are the same else { num = ties * (ties - 1) / 2; } if (smax(hard, cur)) { cnt = num; } else if (hard == cur) { cnt += num; } } // processing parent dist & parent count int l1 = 0, l2 = 0, cnt1 = 0, cnt2 = 0; for (auto [k, v] : opt) { // all paths will increase by len 1 if (k + 1 > l1) { swap(l1, l2), swap(cnt1, cnt2); l1 = k + 1, cnt1 = v; } else if (k + 1 == l1) { cnt1 += v; } else if (k + 1 > l2) { l2 = k + 1, cnt2 = v; } else if (k + 1 == l2) { cnt2 += v; } } for (int v : adj[u]) if (v != p) { if (d[v] + 2 == l1) { (c[v] == cnt1) ? dfs2(v, u, l2, cnt2, dfs2) : dfs2(v, u, l1, cnt1 - c[v], dfs2); } else { dfs2(v, u, l1, cnt1, dfs2); } } }; dfs2(0, -1, 0, 1, dfs2); cout << hard << ' ' << cnt << '\n'; }
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