题解 | 第 k 小

区间第k小用主席树，全局第k小用权值线段树即可，离线把数据离散化后用线段树二分找具体位置即可。

#include<bits/stdc++.h>
#define int long long
using namespace std;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m, k;
    cin >> n >> m >> k;
    vector<int> a(n + 1);
    vector<int> b;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        b.push_back(a[i]);
    }

    vector<array<int, 2>> q;
    
    for (int i = 1; i <= m; i++) {
        int op; cin >> op;
        if (op == 1) {
            int x; cin >> x;
            b.push_back(x);
            q.push_back({1, x});
        }
        if (op == 2) q.push_back({2, 0});
    }
    b.push_back(-1);
    sort(b.begin(), b.end());
    b.erase(unique(b.begin(), b.end()), b.end());
    int sz = b.size() + 1;

    vector<int> tr(sz << 2);

    auto update = [&](auto &self, int l, int r, int x, int p) {
        if (l == r) {
            tr[p]++;
            return;
        }
        int mid = l + r >> 1;
        if (x <= mid) self(self, l, mid, x, p << 1);
        else self(self, mid + 1, r, x, p << 1 | 1);
        tr[p] = tr[p << 1] + tr[p << 1 | 1];
    };

    auto query = [&](auto &self, int l, int r, int k, int p) {
        if (l == r) return l;
        int mid = l + r >> 1;
        int sum = tr[p << 1];
        if (sum >= k) return self(self, l, mid, k, p << 1);
        else return self(self, mid + 1, r, k - sum, p << 1 | 1);
    };

    for (int i = 1; i <= n; i++) {
        int x = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
        update(update, 1, sz, x, 1);
    }

    for (int i = 0; i < m; i++) {
        int op = q[i][0];
        if (op == 1) {
            int x = lower_bound(b.begin(), b.end(), q[i][1]) - b.begin();
            update(update, 1, sz, x, 1);
        }
        else {
            if (tr[1] < k) cout << -1 << "\n";
            else cout << b[query(query, 1, sz, k, 1)] << "\n";
        }
    }


    return 0;
}