Submission #362902

#	Time	Username	Problem	Language	Result	Execution time	Memory
362902		ACmachine	Sailing Race (CEOI12_race)	C++17	50 / 100	3081 ms	4460 KiB

race

#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;
}

• Subtask 1: 50 / 100