CF1227D Optimal Subsequences

思路：

首先对于单个查询(k, p)来说，答案一定是a数组中的前k大数。如果第k大的数字有多个怎么办？取索引最小的若干个。所以我们只需对a数组按照值降序，索引升序排序即可。

多个查询怎么办？离线处理。依然是对数组a按照上面的方法排序，然后对所有的查询按照k升序排序，分别离线计算。具体做法是，按顺序扫描a，用一个集合动态维护前k大的数，当发现某个查询正好询问前k大时，即时处理即可。

那么如何从一个集合中快速找到第k大数？c++标准库中原始的set是不行的，可以用1. 二分+树状数组（或线段树）2. __gnu_pbds https://www.geeksforgeeks.org/ordered-set-gnu-c-pbds/

实现1（二分+树状数组，n * log(n) * log(n)）：

#include <bits/stdc++.h>
 
using namespace std;
const int MAXN = 200000;
typedef pair<int, int> pii;
 
int a[MAXN + 5], bit[MAXN + 5];
 
int lowbit(int x) { return x & -x; }
 
void add(int i, int x)
{
    while (i <= MAXN) { bit[i] += x; i += lowbit(i); }
}
 
int sum(int i)
{
    int ans = 0;
    while (i) { ans += bit[i]; i -= lowbit(i); }
    return ans;
}
 
bool cmp(pii& a, pii& b)
{
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second; 
}
 
void work(pii& q, map<pii, int>& mp)
{
    int k = q.first, p = q.second;
    int l = 1, r = MAXN, res = -1;
    while (l <= r)
    {
        int m = l + r >> 1;
        if (sum(m) < p) l = m + 1;
        else { res = m; r = m - 1; }
    }
    assert(res != -1);
    mp[q] = res;
}
 
int main()
{
    ios::sync_with_stdio(false);
    int n, m;
    while (cin >> n)
    {
        memset(bit, 0, sizeof bit);
        vector<int> a(n);
        for (int i = 0; i < n; i++) cin >> a[i];
        vector<pii> b(n);
        for (int i = 0; i < n; i++) { b[i].first = a[i]; b[i].second = i + 1; }
        sort(b.begin(), b.end(), cmp);
        cin >> m;
        vector<pii> q(m);
        for (int i = 0; i < m; i++) cin >> q[i].first >> q[i].second;
        vector<pii> sorted_q(q.begin(), q.end());
        sort(sorted_q.begin(), sorted_q.end());
        map<pii, int> res;
        int cur = 0;
        for (int i = 0; i < n; i++)
        {
            add(b[i].second, 1);
            while (cur < m && i + 1 == sorted_q[cur].first)
            {
                work(sorted_q[cur], res);
                cur++;
            }
        }
        for (auto it: q)
        {
            int id = res[it];
            cout << a[id - 1] << endl;
        }
    }
    return 0;
}

实现2（__gnu_pbds，n * log(n)）:

#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;
 
typedef pair<int, int> pii;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
 
bool cmp(pii& a, pii& b)
{
    if (a.first != b.first) return a.first > b.first;
    return a.second < b.second; 
}
 
int main()
{
    ios::sync_with_stdio(false);
    int n, m;
    while (cin >> n)
    {
        vector<int> a(n);
        for (int i = 0; i < n; i++) cin >> a[i];
        vector<pii> b(n);
        for (int i = 0; i < n; i++) { b[i].first = a[i]; b[i].second = i + 1; }
        sort(b.begin(), b.end(), cmp);
        cin >> m;
        vector<pii> q(m);
        for (int i = 0; i < m; i++) cin >> q[i].first >> q[i].second;
        vector<pii> sorted_q(q.begin(), q.end());
        sort(sorted_q.begin(), sorted_q.end());
        map<pii, int> res;
        int cur = 0;
        ordered_set st;
        for (int i = 0; i < n; i++)
        {
            st.insert(b[i].second);
            while (cur < m && i + 1 == sorted_q[cur].first)
            {
                res[sorted_q[cur]] = *st.find_by_order(sorted_q[cur].second - 1);
                cur++;
            }
        }
        for (auto it: q)
        {
            cout << a[res[it] - 1] << endl;
        }
    }
}