Submission #786963

#	Time	Username	Problem	Language	Result	Execution time	Memory
786963		GrindMachine	Tenis (COI19_tenis)	C++17	51 / 100	1080 ms	7828 KiB

tenis

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

remember reading the edi & seeing other people's solutions a long time ago
don't exactly remember the idea, but graph/scc ideas were used for proof? (idk)

winners form a pref in each of the courts
find the smallest pref s.t the multiset of guys in all 3 courts that belong to this pref is the same

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

ll a[5][N], pos[5][N];

void solve(int test_case)
{
    ll n,q; cin >> n >> q;
    rep1(p,3){
        rep1(i,n) cin >> a[p][i];
        rep1(i,n) pos[p][a[p][i]] = i;
    }

    vector<bool> ok(n+5);

    auto upd = [&](){
        ll mn2 = inf2, mx2 = -inf2;
        ll mn3 = inf2, mx3 = -inf2;
        fill(all(ok),0);

        rep1(i,n){
            ll x = a[1][i];
            ok[x] = 1;
            amin(mn2,pos[2][x]);
            amax(mx2,pos[2][x]);
            amin(mn3,pos[3][x]);
            amax(mx3,pos[3][x]);
            if(mn2 == 1 and mx2 == i and mn3 == 1 and mx3 == i){
                break;
            }
        }
    };

    upd();

    while(q--){
        ll t; cin >> t;
        if(t == 1){
            ll i; cin >> i;
            if(ok[i]) cout << "DA" << endl;
            else cout << "NE" << endl;
        }
        else{
            ll p,x,y; cin >> p >> x >> y;
            ll i = pos[p][x], j = pos[p][y];
            swap(a[p][i],a[p][j]);
            swap(pos[p][x],pos[p][y]);
            upd();
        }
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

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