# UVa 231 – Testing the CATCHER

## 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 &lt;bits/stdc++.h&gt;

#define pii             pair &lt;int,int&gt;
#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&lt;&lt;#x &quot; = &quot;&lt;&lt;(x)&lt;&lt;endl
#define VI              vector &lt;int&gt;
#define DBG             pf(&quot;Hi\n&quot;)
#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(&quot;%d&quot;,&amp;a)
#define sfl(a)          scanf(&quot;%lld&quot;,&amp;a)
#define sff(a,b)        scanf(&quot;%d %d&quot;,&amp;a,&amp;b)
#define sffl(a,b)       scanf(&quot;%lld %lld&quot;,&amp;a,&amp;b)
#define sfff(a,b,c)     scanf(&quot;%d %d %d&quot;,&amp;a,&amp;b,&amp;c)
#define sfffl(a,b,c)    scanf(&quot;%lld %lld %lld&quot;,&amp;a,&amp;b,&amp;c)
#define loop(i,n)       for(int i=0;i&lt;n;i++)
#define REP(i,a,b)      for(int i=a;i&lt;b;i++)
#define TEST_CASE(t)    for(int z=1;z&lt;=t;z++)
#define PRINT_CASE      printf(&quot;Case %d: &quot;,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
/*------------------------------------------------*/

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

vector&lt;int&gt;v;
int dp[100005];

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

if(ret!=-1) return ret;

int maxi=0;

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

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

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

Another Iterative solution of this problem is here —

```
#include &lt;bits/stdc++.h&gt;

#define pii pair &lt;int,int&gt;
#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&lt;&lt;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(&quot;%d&quot;,&amp;a)
#define loop(i,a) for(int i=0;i&lt;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(&quot;in.txt&quot;,&quot;r&quot;,stdin);
///freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);
int n;
vector&lt;int&gt;v;
int z=0;
while(cin&gt;&gt;n &amp;&amp; n!=-1)
{
if(z)
cout&lt;&lt;endl;
v.pb(100000);
v.pb(n);
int num;
while(cin&gt;&gt;num &amp;&amp;num!=-1)
v.pb(num);
int sz=SZ(v);
for(int i=0;i&lt;=sz;i++)
L[i]=1;
int ans=1;
for(int i=1;i&lt;sz;i++)
{
for(int j=i+1;j&lt;sz;j++)
if(v[j]&lt;v[i] &amp;&amp; L[i]+1 &gt; L[j])
ans=max(ans,++L[j]);
}
pf(&quot;Test #%d:\n&quot;,++z);
cout&lt;&lt;&quot;  maximum possible interceptions: &quot;&lt;&lt;ans&lt;&lt;endl;
v.clear();
}
return 0;
}

```