제출 #804102

#제출 시각아이디문제언어결과실행 시간메모리
804102arush_aguEvacuation plan (IZhO18_plan)C++17
100 / 100
1238 ms79472 KiB
#include <algorithm> #include <cctype> #include <cmath> #include <cstdio> #include <cstring> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <functional> #include <iomanip> #include <iostream> #include <iterator> #include <list> #include <map> #include <numeric> #include <queue> #include <random> #include <set> #include <sstream> #include <stack> #include <string> #include <unordered_map> #include <unordered_set> #include <vector> #ifdef DEBUG #include <time.h> #endif #define all(a) (a).begin(), (a).end() #define rev(a) (a).rbegin(), (a).rend() #define F first #define S second int recur_depth = 0; #ifdef DEBUG #define dbg(x) \ { \ ++recur_depth; \ auto x_ = x; \ --recur_depth; \ cerr << string(recur_depth, '\t') << "\e[91m" << __func__ << ":" \ << __LINE__ << "\t" << #x << " = " << x_ << "\e[39m" << endl; \ } #else #define dbg(x) #endif using namespace std; using namespace __gnu_pbds; typedef pair<int, int> ii; typedef long long ll; typedef long double ld; typedef pair<ll, ll> llll; typedef vector<int> vi; typedef vector<vi> vvi; typedef vector<pair<int, int>> vii; typedef vector<vii> vvii; typedef vector<ll> vl; typedef vector<vl> vvl; typedef vector<pair<ll, ll>> vll; typedef vector<vll> vvll; typedef vector<bool> vb; template <class type1> using ordered_set = tree<type1, null_type, less<type1>, rb_tree_tag, tree_order_statistics_node_update>; template <typename A, typename B> ostream &operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; } template <typename T_container, typename T = typename enable_if< !is_same<T_container, string>::value, typename T_container::value_type>::type> ostream &operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; } const ll MOD = 1e9 + 7; // const ll MOD = 998244353; const int INF = 1e9; const ld EPS = 1e-9; struct DSU { int n; vi par, sz; DSU(int n) : n(n) { par = vi(n, -1); sz = vi(n, 1); } int find(int x) { if (par[x] == -1) return x; return par[x] = find(par[x]); } void merge(int u, int v) { u = find(u), v = find(v); if (u == v) return; if (sz[u] < sz[v]) swap(u, v); sz[u] += sz[v]; par[v] = u; } bool connected(int u, int v) { return find(u) == find(v); } }; void solve() { int n, m; cin >> n >> m; vvii adj(n); vvi ed(m); for (int i = 0; i < m; i++) { int u, v, w; cin >> u >> v >> w; u--, v--; adj[u].push_back({v, w}), adj[v].push_back({u, w}); ed[i] = {u, v, w}; } int k; cin >> k; vi nuc(k); for (int &x : nuc) cin >> x, x--; vi dist(n, INF); vb vis(n); priority_queue<ii, vii, greater<ii>> pq; for (int &x : nuc) pq.push({0, x}), dist[x] = 0; while (!pq.empty()) { auto [d, u] = pq.top(); pq.pop(); if (vis[u]) continue; vis[u] = 1; for (auto [v, w] : adj[u]) if (!vis[v] && dist[v] > w + d) { dist[v] = d + w; pq.push({d + w, v}); } } sort(all(ed), [&](vi a, vi b) { return min(dist[a[0]], dist[a[1]]) > min(dist[b[0]], dist[b[1]]); }); vi ed_idx; DSU d(n); for (int i = 0; i < m; i++) { int u = ed[i][0], v = ed[i][1]; if (d.connected(u, v)) continue; d.merge(u, v); ed_idx.push_back(i); } vvi nadj(n); for (int &idx : ed_idx) { int u = ed[idx][0], v = ed[idx][1]; nadj[u].push_back(v); nadj[v].push_back(u); } const int MAXLG = 22; vvi par(MAXLG, vi(n, -1)); vi depth(n); function<void(int, int, int)> f = [&](int u, int p, int d) { par[0][u] = p; depth[u] = d; for (int &v : nadj[u]) if (v != p) { f(v, u, d + 1); } }; f(0, -1, 0); for (int j = 1; j < MAXLG; j++) for (int i = 0; i < n; i++) if (par[j - 1][i] != -1) par[j][i] = par[j - 1][par[j - 1][i]]; vvi cost(MAXLG, vi(n)); for (int i = 0; i < n; i++) cost[0][i] = min(dist[i], par[0][i] == -1 ? INF : dist[par[0][i]]); for (int j = 1; j < MAXLG; j++) for (int i = 0; i < n; i++) cost[j][i] = min(cost[j - 1][i], par[j - 1][i] == -1 ? INF : cost[j - 1][par[j - 1][i]]); auto kth = [&](int u, int k) { if (depth[u] < k) return -1; for (int i = MAXLG - 1; i >= 0; i--) if (k & (1 << i)) u = par[i][u]; return u; }; auto lca = [&](int u, int v) { if (depth[u] < depth[v]) swap(u, v); u = kth(u, depth[u] - depth[v]); if (u == v) return u; for (int i = MAXLG - 1; i >= 0; i--) if (par[i][u] != par[i][v]) { u = par[i][u]; v = par[i][v]; } return par[0][u]; }; auto ans = [&](int u, int v) { int l = lca(u, v); int res = INF; int k1 = depth[u] - depth[l]; for (int i = MAXLG - 1; i >= 0; i--) if (k1 & (1 << i)) { res = min(res, cost[i][u]); u = par[i][u]; } int k2 = depth[v] - depth[l]; for (int i = MAXLG - 1; i >= 0; i--) if (k2 & (1 << i)) { res = min(res, cost[i][v]); v = par[i][v]; } return res; }; int q; cin >> q; while (q--) { int s, t; cin >> s >> t; s--, t--; cout << ans(s, t) << "\n"; } } int main() { ios_base::sync_with_stdio(0); cin.tie(NULL); clock_t start = clock(); int test_cases = 1; // cin >> test_cases; while (test_cases--) solve(); #ifdef DEBUG cerr << fixed << setprecision(10) << "\nTime Taken: " << (double)(clock() - start) / CLOCKS_PER_SEC << "s\n"; #endif return 0; }

컴파일 시 표준 에러 (stderr) 메시지

plan.cpp: In function 'int main()':
plan.cpp:266:11: warning: unused variable 'start' [-Wunused-variable]
  266 |   clock_t start = clock();
      |           ^~~~~
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