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
