Submission #669820

#TimeUsernameProblemLanguageResultExecution timeMemory
669820ghostwriterPaths (BOI18_paths)C++17
100 / 100
324 ms97176 KiB
#include <bits/stdc++.h> using namespace std; #ifdef LOCAL #include <debug.h> #else #define debug(...) #endif #define ft front #define bk back #define st first #define nd second #define ins insert #define ers erase #define pb push_back #define pf push_front #define _pb pop_back #define _pf pop_front #define lb lower_bound #define ub upper_bound #define mtp make_tuple #define bg begin #define ed end #define all(x) (x).bg(), (x).ed() #define sz(x) (int)(x).size() typedef long long ll; typedef unsigned long long ull; typedef double db; typedef long double ldb; typedef pair<int, int> pi; typedef pair<ll, ll> pll; typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll; typedef string str; #define FOR(i, l, r) for (int i = (l); i <= (r); ++i) #define FOS(i, r, l) for (int i = (r); i >= (l); --i) #define FRN(i, n) for (int i = 0; i < (n); ++i) #define FSN(i, n) for (int i = (n) - 1; i >= 0; --i) #define EACH(i, x) for (auto &i : (x)) #define WHILE while template<typename T> T gcd(T a, T b) { WHILE(b) { a = a % b; swap(a, b); } return a; } template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; } #define file "TEST" mt19937 rd(chrono::steady_clock::now().time_since_epoch().count()); ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); } /* ---------------------------------------------------------------- END OF TEMPLATE ---------------------------------------------------------------- Tran The Bao - ghostwriter Training for VOI23 gold medal ---------------------------------------------------------------- GOAT ---------------------------------------------------------------- */ const int N = 3e5 + 5; int n, m, k, c[N]; ll d[N][1 << 5], rs = 0; vi adj[N]; void input() { cin >> n >> m >> k; FOR(i, 1, n) { cin >> c[i]; --c[i]; } FOR(i, 1, m) { int u, v; cin >> u >> v; adj[u].pb(v); adj[v].pb(u); } } void solve() { FSN(j, 1 << k) FOR(i, 1, n) { if (!(j & (1 << c[i]))) continue; EACH(z, adj[i]) if (!(j & (1 << c[z]))) d[i][j] += d[z][j | (1 << c[z])] + 1; } FOR(i, 1, n) rs += d[i][1 << c[i]]; cout << rs; } signed main() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); // freopen(file".inp", "r", stdin); // freopen(file".out", "w", stdout); input(); solve(); return 0; } /* ---------------------------------------------------------------- From Benq: stuff you should look for * int overflow, array bounds * special cases (n=1?) * do smth instead of nothing and stay organized * WRITE STUFF DOWN * DON'T GET STUCK ON ONE APPROACH ---------------------------------------------------------------- */
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