Submission #613975

#	Time	Username	Problem	Language	Result	Execution time	Memory
613975		HediChehaidar	Monochrome Points (JOI20_monochrome)	C++17	4 / 100	19 ms	340 KiB

monochrome

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (PGCD)
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} // least common multiple (PPCM)
#define ss second
#define ff first
#define all(x) (x).begin() , (x).end()
#define pb push_back
#define vi vector<int>
#define vii vector<pair<int,int>>
#define vl vector<ll>
#define vll vector<pair<ll,ll>>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pdd  pair<double,double>
#define vdd  vector<pdd>
#define dte  tuple<double , double , double>
using namespace std;
const int INF = 1000*1000*1000; // 1 e 9
const int MOD = 1e9 + 7;//998244353 ;
const double EPS = 0.000000001; // 1 e -9
const ll inf = (ll)1e18;

int n;
string s;
vi v[2];
int st[(int)2e6 + 10];
void update(int pos , int l , int r , int i){
    if(i > r || i  < l) return ;
    if(l == r){
        st[pos] = 1; return ;
    }
    update(2 * pos + 1 , l , (l+r) / 2 , i); update(2 * pos + 2 , (l+r) / 2 + 1 , r, i );
    st[pos] = st[2 * pos + 1] + st[2 * pos + 2];
}
int rq(int pos , int l , int r , int i  , int j){
    if(i > r || j < l) return 0;
    if(i <= l && j >= r) return st[pos];
    return rq(2 * pos + 1 , l , (l+r) / 2 , i , j) + rq(2 * pos + 2 , (l+r) / 2 + 1 , r , i , j);
}

int main(){
    //ifstream fin ("testing.txt");
    //ofstream fout ("output.txt");
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin>>n>>s;
    if(n > 2020) return 0;
    /*for(int i = n ; i < 2 * n ; i++){
        int id = s[i] == 'B';
        v[id].pb(i);
    }
    reverse(all(v[0]));
    reverse(all(v[1]));*/
    ll res = 0;
    for(int j = 0 ; j < n ; j++){
        for(int i = 0 ; i < 7 * n ; i++) st[i] = 0;
        int x = j + n;
        v[0].clear(); v[1].clear();
        for(int i = 0 ; i < n ; i++){
            v[s[x] == 'B'].pb(x);
            x++;
            if(x == 2 * n) x = 0;
        }
        reverse(all(v[0])); reverse(all(v[1]));
        ll ans = 0;
        for(int i = j ; i < j + n ; i++){
            int id = s[i] == 'W';
            if(v[id].back() < j){
                ans += rq(0 , 0 , 2 * n - 1 , j + n , 2 * n - 1);
                ans += rq(0 , 0 , 2 * n - 1 , 0 , v[id].back());
            }
            else ans += rq(0 , 0 , 2 * n - 1 , j + n , v[id].back());
            update(0 , 0 , 2 * n - 1 , v[id].back());
            v[id].pop_back();
        }
        //cout << ans << " ";
        res = max(res , ans);
    }
    //cout << "\n";
    cout << res << "\n";
    return 0;
}

/*
    Think of : BS / DFS / BFS / SSSP / SCC / MSP / MAX FLOW / TOPSORT / LCA / MATRIX / DP(bitmask) / 2 POINTERS / SEG TREE / MATH / UN FIND / MO / HLD
    Read the statement CAREFULLY !!
    Make a GREADY APPROACH !!!! (start from highest / lowest)
    Make your own TESTS !!
    Be careful from CORNER CASES !
*/

• Subtask 1: 4 / 4
• Subtask 2: 0 / 21
• Subtask 3: 0 / 10
• Subtask 4: 0 / 65