Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=422

Solution Idea:

This is a LIS problem. But in this problem input size/number of elements in the list is not mentioned. So we have to use faster solution like NlogK algorithm because if the input list is large then O(n^2) solution can give a TLE. So in this problem we use LIS NlogK approach. We need to just run the algorithm and print the solution.

/*
         +-+ +-+ +-+
         |R| |.| |S|
         +-+ +-+ +-+
*/

#include <bits/stdc++.h>

#define pii             pair <int,int>
#define sc              scanf
#define pf              printf
#define Pi              2*acos(0.0)
#define ms(a,b)         memset(a, b, sizeof(a))
#define pb(a)           push_back(a)
#define MP              make_pair
#define db              double
#define ll              long long
#define EPS             10E-10
#define ff              first
#define ss              second
#define sqr(x)          (x)*(x)
#define D(x)            cout<<#x " = "<<(x)<<endl
#define VI              vector <int>
#define DBG             pf("Hi\n")
#define MOD             100007
#define MAX             10000
#define CIN             ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)           (int)a.size()
#define sf(a)           scanf("%d",&a)
#define sfl(a)          scanf("%lld",&a)
#define sff(a,b)        scanf("%d %d",&a,&b)
#define sffl(a,b)       scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)     scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)    scanf("%lld %lld %lld",&a,&b,&c)
#define loop(i,n)       for(int i=0;i<n;i++)
#define REP(i,a,b)      for(int i=a;i<b;i++)
#define TEST_CASE(t)    for(int z=1;z<=t;z++)
#define PRINT_CASE      printf("Case %d: ",z)
#define all(a)          a.begin(),a.end()
#define intlim          2147483648
#define inf             1000000
#define ull             unsigned long long

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

vector<int>v,print_ans;
int n=0;

int L[1000000];
int I[1000000];

int LIS_NlogK()
{
    loop(i,n+2) I[i]=inf;
    I[0]=-inf;

    int length=0;

    for(int i=1;i<=n;i++)
    {
        int low=0, hi=length, mid;

        while(low<=hi)
        {
            mid=(low+hi)/2;
            if(v[i]>I[mid])
                low=mid+1;
            else
                hi=mid-1;
        }

        I[low]=v[i];
        L[i]=low;
        length=max(length,low);
    }
    return length;
}


int main()
{
//    freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);

    v.pb(-100000);

    int a;
    while(sf(a)==1)
    {
        n++;
        v.pb(a);
    }

    int ans=LIS_NlogK();

    int i;

    int temp=ans;

    for(i=n;i>0;i--)
    {
        if(L[i]==temp)
        {
            temp--;
            print_ans.pb(v[i]);
            break;
        }
    }

    for(--i;i>0 && temp ;i--)
    {
        if(v[i]<print_ans.back() && L[i]==temp)
        {
            print_ans.pb(v[i]);
            temp--;
        }
    }

    printf("%d\n-\n",ans);

    for(i=ans-1;i>=0;i--)
        printf("%d\n",print_ans[i]);

    return 0;
}