poj2533

Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18511 Accepted: 7963 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

 

DP: LIS, poj1836是他的升级版

#include <iostream>
#include <cstring>

using namespace std;

const int Size=1000;
int dp[Size+1];
int sq[Size+1];

int main()
{
    int n;
    int rlt;

    while(cin>>n)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)
        {
            cin>>sq[i];
        }
        //dp
        for(int i=1; i<=n; i++)
        {
            dp[i]=1;
            for(int j=1; j<=i-1; j++)
            {
                if(sq[j]<sq[i])
                {
                    if(dp[i]<dp[j]+1)
                    {
                        dp[i]=dp[j]+1;
                    }
                }
            }
        }
        rlt=0;
        for(int i=1; i<=n; i++)
        {
             if(rlt<dp[i])
                 rlt=dp[i];
        }
        cout<<rlt<<endl;
    }
    return 0;
}