제출 #252814

#제출 시각아이디문제언어결과실행 시간메모리
252814kartelDrzava (COCI15_drzava)C++14
32 / 160
530 ms65540 KiB
#include <bits/stdc++.h> //#include <ext/pb_ds/assoc_container.hpp> //#include <ext/pb_ds/tree_policy.hpp> #define in(x) freopen(x, "r", stdin) #define out(x) freopen(x, "w", stdout) //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("Ofast,nso-stack-protector,unroll-loops,fast-math,-O3") //#define F first //#define S second #define pb push_back #define N +100500 #define M ll(1e9 + 7) #define sz(x) (int)x.size() #define re return #define oo ll(1e9) #define el '\n' #define Max_A int(1e9) //#define el endl #define pii pair <int, int> #define err ld(1e-6) #define Max_S int(3e6) #define last(x) x.back() #define all(x) (x).begin(), (x).end() #define arr_all(x, n) (x + 1), (x + 1 + n) using namespace std; //using namespace __gnu_pbds; //typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set; typedef long long ll; typedef long double ld; ld ans = 1e18; int x[N], y[N], k, n, i, cnt[N]; bool f[35]; vector <int> g[N]; bool mrk[N]; ld rast(int i, int j) {return (sqrt(pow(x[i] - x[j], 2) + pow(y[i] - y[j], 2)));} void dfs(int v) { if (mrk[v]) return; mrk[v] = 1; for (auto u : g[v]) { if (mrk[u]) continue; bool F[35]{0}; for (int i = 0; i < k; i++) F[i] = f[i]; for (int i = 0; i < k; i++) { if (F[i]) f[(i + cnt[u]) % k] = 1; } dfs(u); } } bool gd(ld D) { for (i = 1; i <= n; i++) g[i].clear(), mrk[i] = 0; for (i = 1; i <= n; i++) for (int j = i + 1; j <= n; j++) if (rast(i, j) + err < D) g[i].pb(j), g[j].pb(i); for (i = 1; i <= n; i++) { if (mrk[i]) continue; for (int j = 0; j < k; j++) f[j] = 0; f[cnt[i] % k] = 1; dfs(i); if (f[0]) return 1; } return 0; } int main() { cout.precision(3); cout << fixed; srand(time(0)); ios_base::sync_with_stdio(0); iostream::sync_with_stdio(0); ios::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL); // in("input.txt"); // out("output.txt"); cin >> n >> k; for (i = 1; i <= n; i++) cin >> x[i] >> y[i] >> cnt[i]; ld l = 0; ld r = 1e18; while (l + err < r) { ld md = (l + r + err) / 2.0; if (gd(md)) r = md - err; else l = md; } cout << l; }
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