제출 #233532

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
233532		ne4eHbKa	Tenis (COI19_tenis)	C++17	100 / 100	249 ms	12348 KiB

tenis

//{ <defines>
#include <bits/stdc++.h>
using namespace std;

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,-O3")

#define fr(i, n) for(int i = 0; i < n; ++i)
#define fo(n) fr(i, n)

#define re return
#define ef else if
#define ifn(x) if(!(x))
#define _  << ' ' <<

#define ft first
#define sd second
#define ve vector
#define pb push_back
#define eb emplace_back

#define sz(x) int((x).size())
#define ip2(x) (1 << (x))
#define lp2(x) (1ll << (x))
#define bnd(x) x.begin(), x.end()
#define clr(x, y) memset((x), (y), sizeof (x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef ve<int> vi;

inline ll time() {re chrono :: system_clock().now().time_since_epoch().count();}
mt19937 rnd(time());
mt19937_64 RND(time());

template<typename t> inline void umin(t &a, t b) {a = min(a, b);}
template<typename t> inline void umax(t &a, t b) {a = max(a, b);}

int md = 1e9 + 7;

inline int m_add(int&a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_sum(int a, int b) {a += b; if(a < 0) a += md; if(a >= md) a -= md; re a;}
inline int m_mul(int&a, int b) {re a = 1ll * a * b % md;}
inline int m_prod(int a, int b) {re 1ll * a * b % md;}
inline int m_doub(int a) {a <<= 1; if(a >= md) a -= md; re a;}

int m_bpow(ll A, ll b) {
    int a = A % md;
    ll ans = 1;
    for(ll p = lp2(63 - __builtin_clzll(b)); p; p >>= 1) {
        (ans *= ans) %= md;
        if(p & b)
            (ans *= a) %= md;
    }
    re ans;
}

//const ld pi = arg(complex<ld>(-1, 0));
//const ld pi2 = pi + pi;
const int oo = 2e9;
const ll OO = 4e18;
//} </defines>
const int N = 1e5 + 5;

int n, q;
int f[3][N];

pii mx(pii a, pii b) {re a.ft >= b.ft ? a : b;}

struct tree {
    tree *l, *r;
    pii v;
    int p;
    tree(int tl = 0, int tr = n - 1): l(r = 0), v(tl + 1, tl), p(0) {
        if(tl < tr) {
            int tm = (tl + tr) >> 1;
            l = new tree(tl, tm);
            r = new tree(tm + 1, tr);
            v = mx(l->v, r->v);
        }
    }
    void push() {
        if(p) {
            if(l) l->p += p, r->p += p;
            v.ft += p;
            p = 0;
        }
    }
    void add(int ad, int ql, int qr, int tl = 0, int tr = n - 1) {
        if(ql > qr) re push();
        if(ql == tl && qr == tr) {p -= ad; re push();}
        push();
        int tm = (tl + tr) >> 1;
        l->add(ad, ql, min(qr, tm), tl, tm);
        r->add(ad, max(tm + 1, ql), qr, tm + 1, tr);
        v = mx(l->v, r->v);
    }
    void mod(int i, int s) {
        int t = min(min(f[0][i], f[1][i]), f[2][i]);
        add(s, t, n - 1);
    }
};

int main() {
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    
    cin >> n >> q;
    fo(3) fr(j, n) {
        int v; cin >> v;
        f[i][v - 1] = j;
    }

    tree t {};

    fo(n) t.mod(i, +1);

    fo(q) {
        char tp; cin >> tp;
        if(tp == '1') {
            int x; cin >> x;
            cout << (f[0][x - 1] <= t.v.sd ? "DA\n" : "NE\n");
        } else {
            int p, a, b;
            cin >> p >> a >> b;
            --p, --a, --b;
            t.mod(a, -1);
            t.mod(b, -1);
            swap(f[p][a], f[p][b]);
            t.mod(a, +1);
            t.mod(b, +1);
        }
    }
    
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 12 / 12
• Subtask 4: 21 / 21
• Subtask 5: 49 / 49