[DP] POJ-2533 Longest Ordered Subsequence

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 55238 Accepted: 24770

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

题意:给n个数,求出最长不下降子序列

//已AC代码
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>

using namespace std;

int a[10880];
int dp[10880];

int main()
{
int n;

while(scanf("%d",&n)!=EOF)
{

memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
int keep=0;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
for(int i=1;i<=n;i++)
{
dp[i]=1;
for(int j=1;j<=i;j++)
{
if(a[i]>a[j])
{
dp[i]=max(dp[i],dp[j]+1);

}
}
}
for (int i=1;i<=n;i++)
{
keep=max(keep,dp[i]);
}

cout<<keep<<endl;
}
return 0;
}