Submission #767102

#	Time	Username	Problem	Language	Result	Execution time	Memory
767102		Nelt	Exhibition (JOI19_ho_t2)	C++17	100 / 100	50 ms	5588 KiB

joi2019_ho_t2

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,fma")
 
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
/* DEFINES */
#define S second
#define F first
#define ll long long
#define ull unsigned long long
#define ld long double
#define npos ULLONG_MAX
#define INF LLONG_MAX
#define vv(a) vector<a>
#define pp(a, b) pair<a, b>
#define pq(a) priority_queue<a>
#define qq(a) queue<a>
#define ss(a) set<a>
#define mm(a, b) map<a, b>
#define ump(a, b) unordered_map<a, b>
#define ANDROID                   \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define elif else if
#define endl "\n"
#define allc(a) begin(a), end(a)
#define all(a) a, a + sizeof(a) / sizeof(a[0])
#define pb push_back
#define logi(a) __lg(a)
#define sqrt(a) sqrtl(a)
#define mpr make_pair
#define ins insert
using namespace std;
using namespace __gnu_pbds;
using namespace __cxx11;
typedef char chr;
typedef basic_string<chr> str;
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
void solve()
{
    ll n, m;
    cin >> n >> m;
    pp(ll, ll) a[n];
    ll c[m];
    for (ll i = 0; i < n; i++)
        cin >> a[i].S >> a[i].F;
    for (ll i = 0; i < m; i++) 
        cin >> c[i];
    sort(all(a));
    sort(all(c));
    ll ans = 0, pos;
    for (ll i = n - 1; i >= 0; i--)
    {
        pos = lower_bound(all(c), a[i].S) - c;
        ans += m - pos - 1 >= ans;
    }
    cout << ans << endl;
}
/*
 
*/
int main()
{
    ANDROID
    // precomp();
    ll t = 1;
    // cin >> t;
    for (ll i = 1; i <= t; i++)
        // cout << "Case #" << i << ": ",
        solve();
    cerr << "\nTime elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n";
}

• Subtask 1: 10 / 10
• Subtask 2: 40 / 40
• Subtask 3: 50 / 50