Problem Link : https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=167

Solution Idea:

This is a straight forward LIS problem. Just take input and calculate the Longest increasing sub-sequences and print the answer.

/*
         +-+ +-+ +-+
         |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;
int dp[100005];

int LIS(int u)
{
    int &ret=dp[u];

    if(ret!=-1) return ret;

    int maxi=0;

    for(int i=u+1;i<SZ(v);i++)
    {
        if(v[i]<=v[u])
        {
            maxi=max(maxi,LIS(i));
        }
    }
    return ret=1+maxi;
}

int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int a;
    int z=0;
    while(sf(a) && a!=-1)
    {
        if(z) pf("\n");
        v.pb(a);
        while(sf(a) && a!=-1)
        {
            v.pb(a);
        }
        ms(dp,-1);
        int ans=0;

        for(int i=0;i<SZ(v);i++)
            ans=max(ans,LIS(i));
        pf("Test #%d:\n",++z);
        pf("  maximum possible interceptions: %d\n",ans);
        v.clear();
    }
    return 0;
}

Another Iterative solution of this problem is here —

#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 oo 1<<29
#define dd double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define MAX 100100
#define SZ(a) (int)a.size()
#define getint(a) scanf("%d",&a)
#define loop(i,a) for(int i=0;i<a;i++)
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define rtintlim 46340
#define llim 9223372036854775808
#define rtllim 3037000499
#define ull unsigned long long
#define I int
 
using namespace std;
int L[1000000];
int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int n;
    vector<int>v;
    int z=0;
    while(cin>>n && n!=-1)
    {
        if(z)
            cout<<endl;
        v.pb(100000);
        v.pb(n);
        int num;
        while(cin>>num &&num!=-1)
            v.pb(num);
        int sz=SZ(v);
        for(int i=0;i<=sz;i++)
            L[i]=1;
        int ans=1;
        for(int i=1;i<sz;i++)
        {
            for(int j=i+1;j<sz;j++)
                if(v[j]<v[i] && L[i]+1 > L[j])
                    ans=max(ans,++L[j]);
        }
        pf("Test #%d:\n",++z);
        cout<<"  maximum possible interceptions: "<<ans<<endl;
        v.clear();
    }
    return 0;
}