Submission #1187717

#	Time	Username	Problem	Language	Result	Execution time	Memory
1187717		oviyan_gandhi	Examination (JOI19_examination)	C++17	0 / 100	1259 ms	250956 KiB

examination

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define int long long
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using oset = tree<T, null_type, greater_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define N 1000000
struct Score {
    int x, y, z, i;

    bool operator < (const Score &o) const { return make_tuple(z, x, y) > make_tuple(o.z, o.x, o.y); }
} a[N], qu[N];
int ans[N];

struct SegTree {
    int rl, rr, mid;
    SegTree *l, *r;
    oset<int> st;

    SegTree(int rl, int rr) : rl{rl}, rr{rr}, mid{(rl+rr)/2}, l{nullptr}, r{nullptr} {}

    void update(int x, int y) {
        st.insert(y);
        if (rl == rr) return;
        if (x <= mid) {
            if (l == nullptr) l = new SegTree(rl, mid);
            l->update(x, y);
        }
        else {
            if (r == nullptr) r = new SegTree(mid+1, rr);
            r->update(x, y);
        }
    }

    int query(int x, int y) {
        if (rr < x)
            return 0;
        if (x <= rl)
            return st.order_of_key(y-1);
        int ans = 0;
        if (l != nullptr) ans += l->query(x, y);
        if (r != nullptr) ans += r->query(x, y);
        return ans;
    }
};

int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, q; cin >> n >> q;
    for (int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y;
        a[i].z = a[i].x + a[i].y;
    }
    sort(a, a+n);
    for (int i = 0; i < q; i++) {
        cin >> qu[i].x >> qu[i].y >> qu[i].z;
        qu[i].i = i;
    }
    sort(qu, qu+q);
    SegTree seg(1LL, 20000000000LL);
    int i = 0;
    for (int qi = 0; qi < q; qi++) {
        for (; i < n && a[i].z >= qu[qi].z; i++)
            seg.update(a[i].x, a[i].y);
        ans[qu[qi].i] = seg.query(qu[qi].x, qu[qi].y);
    }
    for (int i = 0; i < q; i++)
        cout << ans[i] << '\n';
    return 0;
}

• Subtask 1: 0 / 2
• Subtask 2: 0 / 20
• Subtask 3: 0 / 21
• Subtask 4: 0 / 57