MDOLLS - Nested Dolls spoj , how to solve ?

general
,
1

I can’t understand what question wants to say.
help me to understand this problem.

2

You have to output as minimum number of separate nested dolls as possible like In first test case
3
20 30 40 50 30 40
You see all the dolls can be fit in one doll that is {20,30} doll in {30,40} in {40,50}. so ans is 1 doll. Similarly in second test case
4
20 30 10 10 30 20 40 50
{10,10} in {20,30} and{30,20} in {40,50} (or the {20,30} doll can be fitted in {40,50} and {30,20} separate)
so there need to be 2 different nested dolls. so ans is 2
In third test case no dolls can be fitted inside. So ans is 3
In fourth test case {10,10} in {20,30} in {40,50} and {39,51} is separate. so ans is 2.

3

how LIS will solve this problem ? I can’t figure out/

4

Sort by height and keep extracting lis until you have no element left the number of times you extract is the ans I think as you would want to put as many dolls as possible in each nested doll

5

This code is showing segmentation error but it is not showing error in my compiler
can anyone pls check where there is mistake
#include <bits/stdc++.h>
using namespace std;
void solve();
int main(){
int T;
cin>>T;
while(T–){
solve();
}

return 0;

}
void solve(){
int n,a[10005];
cin>>n;
int p;
p=n;
for(int i=0;i<2n;i++){
cin>>a[i];
}
int x=0,y=0;
x=a[0];
y=a[1];
for(int i=2;i<2n-1;i++){
if(x<a[i] && y<a[i+1]) {
x = a[i];
y = a[i + 1];
p–;
}
else if(x>a[i]&& y>a[i+1]){
p–;
}
i++;
}
cout<<p<<endl;
}