Submission #885349

#	Time	Username	Problem	Language	Result	Execution time	Memory
885349		Koyote	Fortune Telling 2 (JOI14_fortune_telling2)	C++11	100 / 100	615 ms	222580 KiB

fortune_telling2

#include <bits/stdc++.h>
using namespace std;
 
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
 
// Persistent Segment Tree
struct node {
    node *l, *r;
    int val;
    node() : l(0), r(0), val(0) {}
    node(int _v) : l(0), r(0), val(_v) {}
    node(node *_l, node *_r) : l(_l), r(_r), val(l->val + r->val) {}
    void extend() {
        if (!l) l = new node();
        if (!r) r = new node();
    }
};
 
node* update(node *cur, int p, int v, int l, int r) {
    if (l == r) return new node(cur->val + v);
    cur->extend();
    int mid = (l + r) >> 1;
    if (p <= mid) return new node(update(cur->l, p, v, l, mid), cur->r);
    else return new node(cur->l, update(cur->r, p, v, mid + 1, r));
}
 
int query(node *cur, int qs, int qe, int l, int r) {
    if (r < qs || qe < l) return 0;
    if (qs <= l && r <= qe) return cur->val;
    cur->extend();
    int mid = (l + r) >> 1;
    return query(cur->l, qs, qe, l, mid) + query(cur->r, qs, qe, mid + 1, r);
}
// End of Persistent Segment Tree
 
const int N = 2e5 + 7, N2 = 6e5 + 2, LG = 22;
int n, k, a[N], b[N], t[N], t_idx[N2], max_t_idx[LG][N2];
vector<node*> roots; // PST
 
int lg2(int x) { return 31 - __builtin_clz(x); }
 
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    cin >> n >> k, roots.push_back(new node());
    for (int i = 1; i <= n; i++) cin >> a[i] >> b[i];
    for (int i = 1; i <= k; i++) cin >> t[i];
 
    basic_string<int> cmpr; cmpr.reserve(2 * n + k + 1), cmpr += 0;
    for (int i = 1; i <= n; i++) cmpr += a[i], cmpr += b[i];
    for (int i = 1; i <= k; i++) cmpr += t[i];
    sort(all(cmpr)), cmpr.erase(unique(all(cmpr)), cmpr.end());

    auto search = [&](int v) -> int { return lower_bound(all(cmpr), v) - cmpr.begin(); };
    for (int i = 1; i <= n; i++) a[i] = search(a[i]), b[i] = search(b[i]);
    for (int i = 1; i <= k; i++) t[i] = search(t[i]);
 
    // for (int i = 0; i <= sz(cmpr); i++) t_idx[i] = -1;
    for (int i = 1; i <= k; i++) t_idx[t[i]] = i;

    // Sparse table
    for (int i = 1; i <= sz(cmpr); i++)
        max_t_idx[0][i] = t_idx[i];
    for (int j = 1; j < LG; j++)
        for (int i = 1; i + (1 << j) - 1 <= sz(cmpr); i++)
            max_t_idx[j][i] = max(max_t_idx[j - 1][i], max_t_idx[j - 1][i + (1 << (j - 1))]);
    
    auto query_max_t_idx = [&](int l, int r) -> int {
        int len = lg2(r - l + 1);
        return max(max_t_idx[len][l], max_t_idx[len][r - (1 << len) + 1]);
    };
    // End of Sparse table

    // Persistent Segment Tree
    for (int i = 1; i <= k; i++) 
        roots.push_back(update(roots.back(), t[i], 1, 1, sz(cmpr)));
    auto cnt_on_range = [&](int l, int r, int x, int y) {
        // Number of elements a[l..r] that x <= a[i] <= y;
        return query(roots[r], x, y, 1, sz(cmpr)) - query(roots[l - 1], x, y, 1, sz(cmpr));
    };
    // End of Persistent Segment Tree
 
    basic_string<int> ans;
    for (int i = 1; i <= n; i++) {
        if (a[i] == b[i]) { ans += cmpr[a[i]]; continue; }
        bool swapped = (a[i] > b[i] ? (swap(a[i], b[i]), true) : false);
        int last_pos = query_max_t_idx(a[i], b[i] - 1);
        int cnt_flipped = cnt_on_range(last_pos + 1, k, b[i], sz(cmpr));
 
        if (last_pos != 0 || swapped) swap(a[i], b[i]);
        ans += cmpr[~cnt_flipped & 1 ? a[i] : b[i]];
    }
    cout << accumulate(ans.begin(), ans.end(), 0LL) << '\n';
}

• Subtask 1: 4 / 4
• Subtask 2: 31 / 31
• Subtask 3: 65 / 65