[可持久化权值线段树上二分] Just h-index

求区间x,y内的h-index（最大的h有h篇论文引用量不小于h）

解题思路：

在树上x,y区间内二分答案check

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
  
const int MAXN = 2e5 + 10;
  
int fst, lst;
 
int N;
 
struct node { int lson, rson, num;} T[MAXN * 32];
int root[MAXN], cnt;
 
void prt() { cnt = 0; fill(root, root + N + 10, 0);}
 
void update(int l, int r, int &now, int pre, int a)
{
    now = ++cnt;
    T[now] = T[pre];
    T[now].num++;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(a <= mid) update(l, mid, T[now].lson, T[pre].lson, a);
    else update(mid + 1, r, T[now].rson, T[pre].rson, a);
}
 
int sum = 0;
 
bool gao(int l, int r, int now, int pre, int pos)
{
    if(l == r)
    {
        sum += ( T[now].num - T[pre].num );
        return sum >= pos;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid)
    {
        sum += T[T[now].rson].num - T[T[pre].rson].num;
        gao(l, mid, T[now].lson, T[pre].lson, pos);
    }
    else
    {
        gao(mid + 1, r, T[now].rson, T[pre].rson, pos);
    }
}
 
int a[MAXN] = {0};
  
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
 
    int n, m;
     
    while( cin >> n >> m )
    {
        prt();
         
        N = 0;
     
        for(int i = 1; i <= n; i++)
            cin >> a[i], N = max(N, a[i] + 10);
         
        for(int i = 1; i <= n; i++)
            update(0, N, root[i], root[i - 1], a[i]);
         
        int a, b;
         
        for(int i = 1; i <= m; i++)
        {
            cin >> a >> b;
             
            fst = 0, lst = N - 1;
             
            int ans = 0, mid;
             
            while(fst <= lst)
            {
                mid = (fst + lst) / 2;
                sum = 0;
                if(gao(0, N, root[b], root[a - 1], mid))
                    ans = mid, fst = mid + 1;
                else
                    lst = mid - 1;
            }
             
            cout << ans << '
';
             
        }  
    }
     
     
     
    return 0;
}