제출 #233369

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
233369		ne4eHbKa	Tenis (COI19_tenis)	C++17	51 / 100	1094 ms	5240 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 p[3][N], ps[3][N];
bool win[N];
bool ready;

void count() {
    ready = true;
    memset(win, 0, n);
    queue<int> q;
    fo(3) q.push(p[i][0]);
    vi first(3, 0);
    for(; !q.empty(); q.pop()) {
        int v = q.front();
        if(win[v]) continue;
        win[v] = true;
        fo(3) {
            for(int j = first[i]; j < ps[i][v]; ++j) q.push(p[i][j]);
            first[i] = ps[i][v] + 1;
        }
    }
}

void solve() {
    cin >> n >> q;
    fr(j, 3) fo(n) cin >> p[j][i], ps[j][--p[j][i]] = i;
    ready = false;
    fo(q) {
        char tp; cin >> tp;
        if(tp == '1') {
            if(!ready) count();
            int x; cin >> x;
            cout << (win[x - 1] ? "DA\n" : "NE\n");
        } else {
            ready = false;
            int f, a, b;
            cin >> f >> a >> b;
            --f, --a, --b;
            swap(p[f][ps[f][a]], p[f][ps[f][b]]);
            swap(ps[f][a], ps[f][b]);
        }
    }
}

int main() {
#ifdef _LOCAL
    freopen("in.txt", "r", stdin);
    int tests; cin >> tests;
    for(int test = 1; test <= tests; ++test) {
        cerr << test << " {\n";
        solve();
        cerr << "}\n\n";
    }
#else
    //freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
    ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
    solve();
#endif
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 12 / 12
• Subtask 4: 21 / 21
• Subtask 5: 0 / 49