POJ 2533, Longest Ordered Subsequence

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 12078  Accepted: 5098

Description
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.

 

Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000

 

Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

 

Sample Input
7
1 7 3 5 9 4 8

 

Sample Output
4

Source
Northeastern Europe 2002, Far-Eastern Subregion

---

//  POJ2533.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < algorithm >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     int  N;
    scanf( " %d " ,  & N);

     int  num[ 1000 ];
     for  ( int  i  =   0 ; i  <  N;  ++ i)scanf( " %d " ,  & num[i]);

     int  DP[ 1000 ];    
    fill( & DP[ 0 ],  & DP[N],  1 );

     for  ( int  i  =   1 ; i  <  N;  ++ i)
         for  ( int  j  =   0 ; j  <  i;  ++ j)
             if  (num[j]  <  num[i]) DP[i]  =  max(DP[j]  +   1 , DP[i]);

    cout  <<   * max_element( & DP[ 0 ],  & DP[N]) << endl;

     return   0 ;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/13/1582362.html