Submission #220668

#	Time	Username	Problem	Language	Result	Execution time	Memory
220668		srvlt	JJOOII 2 (JOI20_ho_t2)	C++14	100 / 100	16 ms	7180 KiB

ho_t2

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define db long double
#define pb push_back
#define ppb pop_back
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(), (x).end()

void dout() { cerr << '\n'; }

template <typename Head, typename... Tail>
void dout(Head H, Tail... T) {
    cerr << " " << H;
    dout(T...);
}

#ifdef LOCAL
    #define dbg(...) cerr << #__VA_ARGS__, dout(__VA_ARGS__)
#else
    #define dbg(...) ;
#endif

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef pair <int, int> pii;

const int N = 2e5 + 123;
int n, k, p[3][N], sf[3][N];
int l[3][N], r[3][N];

int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL);
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
    #endif

    string s;
    cin >> n >> k >> s;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < 3; j++) {
            p[j][i] = p[j][i - 1];
        }
        if (s[i - 1] == 'J') {
            p[0][i]++;
            l[0][p[0][i]] = i;
        }
        if (s[i - 1] == 'O') {
            p[1][i]++;
            l[1][p[1][i]] = i;
        }
        if (s[i - 1] == 'I') {
            p[2][i]++;
            l[2][p[2][i]] = i;
        }
    }
    for (int i = n; i >= 1; i--) {
        for (int j = 0; j < 3; j++) {
            sf[j][i] = sf[j][i + 1];
        }
        if (s[i - 1] == 'J') {
            sf[0][i]++;
            r[0][sf[0][i]] = i;
        }
        if (s[i - 1] == 'O') {
            sf[1][i]++;
            r[1][sf[1][i]] = i;
        }
        if (s[i - 1] == 'I') {
            sf[2][i]++;
            r[2][sf[2][i]] = i;
        }
    }
    int j = 1, ans = N;
    for (int i = 1; i <= n; i++) {
        while (j < i && p[1][i] - p[1][j] >= k) {
            j++;
        }
        if (s[i - 1] == 'O' && p[1][i] - p[1][j - 1] == k) {
            if (r[0][sf[0][j] + k] && l[2][p[2][i] + k]) {
                ans = min(ans, l[2][p[2][i] + k] - r[0][sf[0][j] + k]);
            }
        }
    }
    if (ans == N) {
        cout << -1;
        return 0;
    }
    cout << ans - 3 * k + 1;
}

• Subtask 1: 1 / 1
• Subtask 2: 12 / 12
• Subtask 3: 87 / 87