Memory Limit Exceeded with Segment Tree in SPOJ Posters? - memory-management

Question

Given a horizontal section of wall , and N layers of paints applied from co-ordinates Xi to Yi , Output the distinct number of layers visible.
Here is the problem link http://www.spoj.com/problems/POSTERS/
Here is my solution http://ideone.com/gBJKnL
Approach :
I tried solving the problem by lazily updating child node values through a Segment Tree , the most recent value replaces the older one in their lazy updates. This way only the recent paint gets applied into the horizontal cross-section. although the code works fine on custom test cases , It takes a lot of memory and gets aborted by the Online Judge .
#include <iostream>
#include <set>
#include <vector>
#define MAX 10000000+100
typedef long long int ll;
using namespace std;
ll Tree[3*MAX],lazy[MAX*2];
void Update(ll s,ll start,ll end,ll left,ll right,ll value)
{
if(lazy[s]!=0)
{
Tree[s]=(lazy[s]*(end-start+1));
if(start!=end)lazy[2*s+1]=lazy[s],lazy[s*2+2]=lazy[s];
lazy[s]=0;
}
if(start>end||left>end||right<start)return;
if(start>=left&&end<=right)
{
Tree[s] = (value*(end-start+1));
if(start!=end)
{
lazy[2*s+1]=value;
lazy[2*s+2]=value;
}
return ;
}
ll mid=(start+end)/2;
Update(2*s+1,start,mid,left,right,value);
Update(2*s+2,mid+1,end,left,right,value);
Tree[s] = Tree[s*2+1]+Tree[2*s+2];
}
ll Read(ll s,ll start,ll end,ll left,ll right)
{
if(start>end||start>right||end<left)return 0;
if(lazy[s]!=0)
{
Tree[s]=(lazy[s]*(end-start+1));
if(start!=end)
{
lazy[2*s+1]=lazy[s];
lazy[2*s+2]=lazy[s];
}
lazy[s]=0;
}
if(start>=left&&end<=right)return Tree[s];
else return (Read(2*s+1,start,(start+end)/2,left,right)+Read(2*s+2,1+((start+end)/2),end,left,right));
}
int main() {
// your code goes here
ll t;
cin>>t;
while(t--)
{
ll n,z=1,li=-1;
cin>>n;
vector<pair<ll,ll> > b;
for(ll i=0;i<n;i++)
{
ll u,v;
li = max(li,v);
cin>>u>>v;
b.push_back(make_pair(u-1,v-1));
}
for(auto v: b)
Update(0,0,li+2,v.first,v.second,z++);
set<ll> a;
for(ll i=0;i<li+2;i++)cout<<Read(0,0,li+2,i,i)<<" ",a.insert(Read(0,0,li+2,i,i));
cout<<endl;
cout<<a.size()-1<<endl;
}
return 0;
}

Answer 1

Here is how you should be doing it:
#include <bits/stdc++.h>
#define mx 400005
using namespace std;
int tr[mx], lz[mx];
int t, n, l, r;
void update(int node, int s, int e, int l, int r, int val)
{
if(lz[node]!=0)
{
tr[node]=lz[node];
if(s!=e)
{
lz[node*2]=lz[node];
lz[node*2+1]=lz[node];
}
lz[node]=0;
}
if(s>e || r<s || l>e)
return;
if(s>=l && e<=r)
{
tr[node]=val;
if(s!=e)
{
lz[2*node]=val;
lz[2*node+1]=val;
}
return;
}
int m=s+(e-s)/2;
update(2*node,s,m,l,r,val);
update(2*node+1,m+1,e,l,r,val);
tr[node]=val;
//tr[node]=max(tr[2*node],tr[2*node+1]);
}
int query(int node, int s, int e, int l, int r)
{
if(r<s || e<l)
return 0;
if(lz[node]!=0)
{
tr[node]=lz[node];
if(s!=e)
{
lz[node*2]=lz[node];
lz[node*2+1]=lz[node];
}
lz[node]=0;
}
if(l<=s && r>=e)
return tr[node];
int m=s+(e-s)/2;
return query(2*node,s,m,l,r)+query(2*node+1,m+1,e,l,r);
}
int main()
{
//cout << "Hello world!" << endl;
cin>>t;
while(t--)
{
for(int i=0; i<mx; i++) tr[i]=0;
cin>>n;
int lr[n+1][2];
map<int,bool> mark;
vector<int> vec;
//int c=0;
for(int i=0; i<n; i++)
{
cin>>l>>r;
lr[i][0]=l;
lr[i][1]=r;
if(mark[l]==0)
{
vec.push_back(l);
mark[l]=1;
}
if(mark[r]==0)
{
vec.push_back(r);
mark[r]=1;
}
}
sort(vec.begin(), vec.end());
map<int,int> mp;
int c=1;
for(int i=0; i<vec.size(); i++)
mp[vec[i]]=c++;
for(int i=0; i<n; i++)
{
//cout<<mp[lr[i][0]]<<" "<<mp[lr[i][1]]<<"\n";
update(1,1,vec.size(),mp[lr[i][0]],mp[lr[i][1]],i+1);
}
set<int> ans;
for(int i=1; i<=vec.size(); i++)
{
//cout<<query(1,1,vec.size(),i,i)<<" ";
ans.insert(query(1,1,vec.size(),i,i));
}
int k = ans.size();
if(ans.find(0) != ans.end())
k--;
printf("%d\n",k);
}
return 0;
}