Submission #362902

#TimeUsernameProblemLanguageResultExecution timeMemory
362902ACmachineSailing Race (CEOI12_race)C++17
50 / 100
3081 ms4460 KiB
#include <bits/stdc++.h> using namespace std; #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long ll; typedef long double ld; typedef double db; typedef string str; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pii> vpii; typedef vector<pll> vpll; typedef vector<str> vstr; #define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in) #define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in) #define REP(i,b) FOR(i,0,b,1) #define REPD(i,b) FORD(i,b,0,1) #define pb push_back #define mp make_pair #define ff first #define ss second #define all(x) begin(x), end(x) #define rsz resize #define MANY_TESTS int tcase; cin >> tcase; while(tcase--) const double EPS = 1e-9; const int MOD = 1e9+7; // 998244353; const ll INFF = 1e18; const int INF = 1e9; const ld PI = acos((ld)-1); const vi dy = {1, 0, -1, 0, -1, 1, 1, -1}; const vi dx = {0, 1, 0, -1, -1, 1, -1, 1}; #ifdef DEBUG #define DBG if(1) #else #define DBG if(0) #endif #define dbg(x) cout << "(" << #x << " : " << x << ")" << endl; // ostreams template <class T, class U> ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;} template <class T> ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template <class T, class U> ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; } template<class T> ostream& operator<<(ostream& out, const vector<T> &v){ out << "["; REP(i, v.size()) out << v[i] << ", "; out << "]"; return out;} // istreams template<class T> istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; } template<class T, class U> istream& operator>>(istream& in, pair<T, U> &p){ in >> p.ff >> p.ss; return in; } //searches template<typename T, typename U> T bsl(T lo, T hi, U f){ hi++; T mid; while(lo < hi){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid+1; } return lo; } template<typename U> double bsld(double lo, double hi, U f, double p = 1e-9){ int r = 3 + (int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? hi = mid : lo = mid; } return (lo + hi)/2; } template<typename T, typename U> T bsh(T lo, T hi, U f){ lo--; T mid; while(lo < hi){ mid = (lo + hi + 1)/2; f(mid) ? lo = mid : hi = mid-1; } return lo; } template<typename U> double bshd(double lo, double hi, U f, double p = 1e-9){ int r = 3+(int)log2((hi - lo)/p); double mid; while(r--){ mid = (lo + hi)/2; f(mid) ? lo = mid : hi = mid; } return (lo + hi)/2; } // some more utility functions template<typename T> pair<T, int> get_min(vector<T> &v){ typename vector<T> :: iterator it = min_element(v.begin(), v.end()); return mp(*it, it - v.begin());} template<typename T> pair<T, int> get_max(vector<T> &v){ typename vector<T> :: iterator it = max_element(v.begin(), v.end()); return mp(*it, it - v.begin());} template<typename T> bool ckmin(T& a, const T& b){return b < a ? a = b , true : false;} template<typename T> bool ckmax(T& a, const T& b){return b > a ? a = b, true : false;} int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n, T; cin >> n >> T; vector<vector<array<int, 2>>> dp(n, vector<array<int, 2>>(n, {0, 0})); vector<vector<bool>> mat(n, vector<bool>(n, false)); REP(i, n){ int x; cin >> x; while(x != 0){ x--; mat[i][x] = true; cin >> x; } } // dp[i][j][d] -> som na i, obluk mozem kratcat j dopredu / dozadu podla d auto calc_dp = [&](){ REP(i, n){ dp[i][0][0] = dp[i][0][1] = 0; } FOR(ln, 1, n, 1){ REP(i, n){ FOR(k, 1, ln + 1, 1){ int nxt = (i + k) % n; if(!mat[i][nxt]) continue; dp[i][ln][0] = max(dp[i][ln][0], 1 + dp[nxt][k - 1][1]); dp[i][ln][0] = max(dp[i][ln][0], 1 + dp[nxt][ln - k][0]); } FOR(k, 1, ln + 1, 1){ int nxt = (i - k); if(nxt < 0) nxt += n; if(!mat[i][nxt]) continue; dp[i][ln][1] = max(dp[i][ln][1], 1 + dp[nxt][k - 1][0]); dp[i][ln][1] = max(dp[i][ln][1], 1 + dp[nxt][ln - k][1]); } } } }; calc_dp(); int maxs_without_intersection = 0; REP(i, n){ if(dp[i][n - 1][0] > dp[maxs_without_intersection][n - 1][0]) maxs_without_intersection = i; } if(T == 0){ cout << dp[maxs_without_intersection][n - 1][0] << "\n"; cout << maxs_without_intersection + 1 << "\n"; return 0; } // ak chcem len raz intersectnut prvy tah; // spravim prvy tah, dostanem 2 obluky; v jednom obluku musim robit zatacky len rovnakym smerom; potom skocim na druhy obluk a transitionujem do hornej dp tabulky; // dp_t[i][j][0] -> maximalny pocet tahov jednym smerom ak sa musim posunut presne o j krokov do smeru; -1 ak sa neda vector<vector<array<int, 2>>> dp_t(n, vector<array<int, 2>>(n, {-1, -1})); auto calc_dpt = [&](){ REP(i, n){ dp_t[i][0][0] = dp_t[i][0][1] = 0; FOR(k, 1, n, 1){ int nxt = (i + k) % n; if(mat[i][nxt]) dp_t[i][k][0] = 1; int nxt2 = (i - k); if(nxt2 < 0) nxt2 += n; if(mat[i][nxt2]) dp_t[i][k][1] = 1; } } FOR(ln, 1, n, 1){ REP(i, n){ FOR(k, 1, ln + 1, 1){ int nxt = (i + k) % n; if(dp_t[i][k][0] == -1) continue; if(dp_t[nxt][ln - k][0] != -1) dp_t[i][ln][0] = max(dp_t[i][ln][0], dp_t[i][k][0] + dp_t[nxt][ln - k][0]); } FOR(k, 1, ln + 1, 1){ int nxt = i - k; if(nxt < 0) nxt += n; if(dp_t[i][k][1] == -1) continue; if(dp_t[nxt][ln - k][1] != -1) dp_t[i][ln][1] = max(dp_t[i][ln][1], dp_t[i][k][1] + dp_t[nxt][ln - k][1]); } } } }; calc_dpt(); // spajam tie obluky; bruteforce zaciatocneho pristavu, bruteforce prveho tahu, bruteforce druheho tahu, bruteforce 3 tahu -> O(n^4); // ak zmazeme bruteforce 2 tahu, dostneme sa na O(n^3); // druhy tah pozostava z viacerych tahov jednym smerom -> dpt; po 3 zvysok je zaciatocne dp; int maxs_intersection = 0; int maxs_intersection_value = 0; REP(i, n){ FOR(k, 1, n, 1){ // clockwise prve tahy ->tahy jednym smerom int d1 = (i + k) % n; FOR(g, 1, (n - 1) - k + 1, 1){ // pokracujem clockwise int d2 = (d1 + g) % n; FOR(l, 1, k, 1){ int d3 = (i + l) % n; // volne z prveho tahu if(!mat[d2][d3]) continue; if(dp_t[d1][g][0] == -1) continue; int re = 1 + dp_t[d1][g][0] + 1 + max(dp[d3][k - l - 1][0], dp[d3][l - 1][1]); if(re > maxs_intersection_value){ maxs_intersection_value = re; maxs_intersection = i; } } } } FOR(k, 1, n, 1){ // counterclockwise tahy; int d1 = (i - k); if(d1 < 0) d1 += n; FOR(g, 1, (n - 1) - k + 1, 1){ int d2 = (d1 - g); if(d2 < 0) d2 += n; FOR(l, 1, k, 1){ int d3 = (i - l); if(d3 < 0) d3 += n; if(!mat[d2][d3]) continue; if(dp_t[d1][g][1] == -1) continue; int re = 1 + dp_t[d1][g][1] + 1 + max(dp[d3][k - l - 1][1], dp[d3][l - 1][0]); if(re > maxs_intersection_value){ maxs_intersection_value = re; maxs_intersection = i; } } } } } if(maxs_intersection_value > dp[maxs_without_intersection][n - 1][0]){ cout << maxs_intersection_value << "\n"; cout << maxs_intersection + 1 << "\n"; } else{ cout << dp[maxs_without_intersection][n - 1][0] << "\n"; cout << maxs_without_intersection + 1 << "\n"; } return 0; }
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