Submission #533816

#	Time	Username	Problem	Language	Result	Execution time	Memory
533816		RaresFelix	Crossing (JOI21_crossing)	C++17	26 / 100	1563 ms	191380 KiB

Main

#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;

#define LT(x) ((x) == 'I' ? 2 : (x) == 'O')
#define CMB(x, y) (((x) == (y)) ? (x) : (3 - ((x) + (y))))
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

#define MN 257171
int n, q;
vector<int> C[10], P;
struct SL {
    vector<int> B;
    int Vl[4 * MN], Eq[4 * MN], Lz[4 * MN];
    void bld(int u, int s, int d) {
        Lz[u] = Vl[u] = Eq[u] = 0;
        if(s == d) {
            Vl[u] = B[s];
            Eq[u] = (B[s] == P[s]);
            return;
        }
        bld(u * 2, s, (s + d) / 2);
        bld(u * 2 + 1, (s + d) / 2 + 1, d);
        Vl[u] = ((Vl[u * 2] == Vl[u * 2 + 1]) ? Vl[u * 2] : -1);
        Eq[u] = Eq[u * 2] && Eq[u * 2 + 1];
    }
    SL(vector<int> E) : B(std::move(E)){
        bld(1, 0, n-1);
    }
    void prp(int u, int s, int d) {
        if(!Lz[u]) return;
        int v = Lz[u] - 3;
        Eq[u] = (Vl[u] == v);
        if(s != d)
            Lz[u * 2] = Lz[u * 2 + 1] = Lz[u];
        Lz[u] = 0;
    }
    void pset(int l, int r, int v, int u, int s, int d) {
        prp(u, s, d);
        if(r < s || d < l) return;
        if(l <= s && d <= r) {
            Lz[u] = v + 3;
            prp(u, s, d);
            return;
        }
        pset(l, r, v, u * 2, s, (s + d) / 2);
        pset(l, r, v, u * 2 + 1, (s + d) / 2 + 1, d);
        Vl[u] = ((Vl[u * 2] == Vl[u * 2 + 1] )? Vl[u * 2] : -1);
        Eq[u] = Eq[u * 2] && Eq[u * 2 + 1];
    }
};
vector<SL> SS;
void af() {
    int ok = 0;
    for(auto &it : SS) ok |= it.Eq[1];
    cout << (ok ? "Yes\n" : "No\n");
}
void rd(vi &C) {
    string w;
    cin >> w;
    for(auto it : w) C.push_back(LT(it));
}
int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    cin >> n;
    for(int i = 0; i < 3; ++i) rd(C[i]);
    int nr = 2;
    for(int i = 0; i < 3; ++i)
        for(int j = i + 1; j < 3; ++j)
            for(int k = !(++nr); k < n; ++k) C[nr].push_back(CMB(C[i][k], C[j][k]));
    for(int i = 0; i < n; ++i) C[6].push_back(CMB(C[0][i], CMB(C[1][i], C[2][i])));
    for(int i = 0; i < n; ++i) C[7].push_back(CMB(C[1][i], CMB(C[0][i], C[2][i])));
    for(int i = 0; i < n; ++i) C[8].push_back(CMB(C[2][i], CMB(C[0][i], C[2][i])));

    cin >> q;
    rd(P);
    for(int i = 0; i < 9; ++i)
        SS.emplace_back(C[i]);
    af();
    for(int i = 1, l, r; i <= q; ++i) {
        char c;
        cin >> l >> r >> c;
        --l, --r;
        for(auto &it : SS)
            it.pset(l, r, LT(c), 1, 0, n - 1);
        af();
    }
    return 0;
}

• Subtask 1: 3 / 3
• Subtask 2: 23 / 23
• Subtask 3: 0 / 23
• Subtask 4: 0 / 51