제출 #1064384

#제출 시각아이디문제언어결과실행 시간메모리
1064384c2zi6열쇠 (IOI21_keys)C++17
37 / 100
3005 ms30432 KiB
#define _USE_MATH_DEFINES #include <bits/stdc++.h> #define ff first #define ss second #define pb push_back #define all(a) (a).begin(), (a).end() #define replr(i, a, b) for (int i = int(a); i <= int(b); ++i) #define reprl(i, a, b) for (int i = int(a); i >= int(b); --i) #define rep(i, n) for (int i = 0; i < int(n); ++i) #define mkp(a, b) make_pair(a, b) using namespace std; typedef long long ll; typedef long double ld; typedef pair<int, int> PII; typedef vector<int> VI; typedef vector<PII> VPI; typedef vector<VI> VVI; typedef vector<VVI> VVVI; typedef vector<VPI> VVPI; typedef pair<ll, ll> PLL; typedef vector<ll> VL; typedef vector<PLL> VPL; typedef vector<VL> VVL; typedef vector<VVL> VVVL; typedef vector<VPL> VVPL; template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;} template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;} #include <ext/pb_ds/assoc_container.hpp> using namespace __gnu_pbds; template<class T> using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; VI find_reachable(VI r, VI U_ARG, VI V_ARG, VI c) { int n = r.size(); int m = U_ARG.size(); VVPI edges(n); /* * edges[x] = vor koxer@ kbacvin * x balninq@ stanaluc heto? */ rep(i, m) { edges[c[i]].pb({U_ARG[i], V_ARG[i]}); edges[c[i]].pb({V_ARG[i], U_ARG[i]}); } VI ans(n); rep(s, n) { VI vis(n); VI key(n); VVI gp(n); queue<int> q; /*cout << "SOLVING FOR " << s << endl;*/ auto addkey = [&](int k) { if (key[k]) return; /*cout << " ADDED NEW KEY " << k << endl;*/ key[k] = true; for (auto[u, v] : edges[k]) { if (vis[u] && !vis[v]) { /*cout << " pushed " << v << " to the queue" << endl;*/ q.push(v); vis[v] = true; } gp[u].pb(v); } }; q.push(s); vis[s] = true; addkey(r[s]); while (q.size()) { int u = q.front(); q.pop(); /*cout << "CURRENTLY AT VERTEX " << u << endl;*/ addkey(r[u]); for (int v : gp[u]) if (!vis[v]) { /*cout << " ADDED VERTEX " << v << endl;*/ q.push(v); vis[v] = true; } } int cnt = 0; rep(i, n) cnt += vis[i]; ans[s] = cnt; /*break;*/ } int mn = 2e9; for (int x : ans) setmin(mn, x); for (int& x : ans) x = (x == mn); return ans; }
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