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(log^3 n)
Update Complexity — O(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
Great blog!
I had read about it on commonlounge's merge sort tree post. But it did not had an implementation. This made my doubts about the implementation very clear. Awesome blog.
You can do it quicker using wavelet tree if all queries can be processed offline.
Build: O((n+q)log(q))
Update and query: O(log^2(q))
Oh okay, didn't know about that. This can be used for online queries! I will check wavelet trees. Do send tutorial links if you have.
https://codeforces.net/blog/entry/52854
Here you can see how to do queries without updates. For updates you have to store cartesian tree in every vertex of wavelet tree.
I have done it using C++ STL: Policy-based data structures.
Policy-based data structures alone were sufficient for this functionality.
Edit: My bad, I thought you meant queries on a fixed range.
No, you can't use policy-based ds to range query for kth minimum. Try implementing mkthnum in spoj using only policy based ds.
There is no implemented tree multiset in STL
try comparator which returns true on equal elements
I can't erase elements with less_equal comparator, e.g. this code output "1"
So updation will not be possible.