Hello, I have implemented the version of merge sort tree that can handle point updates and supports duplicate elements in the tree.

I have done it using C++ STL: Policy based data structures.

There is no implemented tree multiset in STL. So, I have used pair<T,int> as a key where the second element in pair is the time when item has been added, in order to maintain uniqueness of elements.

The time complexity summaries are as follows :

• Build Complexity — O(n log^2 n)

• Query Complexity — O(n log^3 n)

• Update Complexity — O(n log^2 n)

The code queries the kth largest element in a range, similar to MKTHNUM in spoj but with an update part.

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
#define ll long long int
#define lc                      ((n)<<1)
#define rc                      ((n)<<1|1)
#define pb push_back
using namespace std;
using namespace __gnu_pbds;

typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, 
             tree_order_statistics_node_update> pds; 


ll n;
const ll N=463005;
vector <ll> arr(N);
pds mst[N];

void init(ll pp)
{
    for(ll i=0;i<4*pp;i++)
    {
        mst[i].clear();
    }
}

void build(ll n,ll b,ll e)
{
    if(b==e)
    {
        mst[n].insert({arr[b],b});
        return;
    }
    for(ll i=b;i<=e;i++)
        mst[n].insert({arr[i],i});
        
    ll mid=(b+e)/2;
    build(lc,b,mid);
    build(rc,mid+1,e);
}

ll query(ll n,ll b,ll e,ll i,ll j,ll v,ll idx)
{
    if(b>j or e<i)  return 0;
    if(b>=i and e<=j)
    {
        ll k=mst[n].order_of_key({v,idx});
        return k;
    }
    ll mid=(b+e)/2;
    return query(lc,b,mid,i,j,v,idx) + query(rc,mid+1,e,i,j,v,idx);
}

void update(ll n,ll b,ll e,ll i,ll v,ll nw)
{
    if(i<b or e<i)  return;
    if(b==i and e==i)
    {
        mst[n].erase(mst[n].find({v,i}));
        mst[n].insert({nw,i});
        return;
    }
    ll mid=(b+e)/2;
    update(lc,b,mid,i,v,nw);
    update(rc,mid+1,e,i,v,nw);
    mst[n].erase(mst[n].find({v,i}));
    mst[n].insert({nw,i});
}

int main()
{
    
    ll test,i,j,queries;
    scanf("%lld",&test); // number of test cases
    while(test--)
    {
        
        scanf("%lld",&n);  //number of elements in the array
        init(n);        // initializing the policy based tree
        
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&arr[i]);  //scanning the array
        }
        
        build(1,1,n);
        
        cin>>queries;       //number of queries
        while(queries--)
        {
            ll choice;
            scanf("%lld",&choice);      //if choice is 0 -> it represents query, choice 1 -> represents update
            if(choice==0)
            {
                ll l,r,x;
                scanf("%lld %lld %lld",&l,&r,&x);       //  the x-th number in sorted a[l ... r] segment
                
                ll low = mst[1].find_by_order(0)->first, high = mst[1].find_by_order(n-1)->first , mid, ans ;
                
                // binary search to find the x-th number
                
                while(low <= high) 
                {
                    mid = low + high >> 1;
                    ll k = query(1,1,n,l,r,mid,1005);       // i have considered the highest element in the array to be 1000, change according to your program
                    if(k >=x ) 
                    {
                        ans = mid;
                        high = mid-1;
                    }
                    else low = mid+1;
                }
                
                printf("%lld\n",ans);
                
            }
            else
            {               
                // Update the profit of the idx-th shop to x
                
                ll idx,x;
                scanf("%lld %lld",&idx,&x);
                update(1,1,n,idx,arr[idx],x);
                arr[idx]=x;
            }
            
        }
        
    }
    return 0;
    
}

Execution with sample test case : https://ideone.com/kQrfVe

Do comment if you find any mistake.

Thank You