POJ 2533 Longest Ordered Subsequence

 

 

Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 35605   Accepted: 15621

 

Description

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence (a₁, a₂, ..., a_N) be any sequence (a_i1, a_i2, ..., a_iK), where 1 <= i₁ < i₂ < ... < i_K <= 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

題意:求最長上升子序列長度

題解:DP 這題我WA了兩發.....~~~~(>_<)~~~~

AC代碼:

 

#include
#include
#define N 1005
using namespace std;
int dp[N],num[N],n;
int main()
{
    cin.sync_with_stdio(false);
    while(cin>>n){
        int res=1;
        for(int i=0;i>num[i];
        for(int i=1;inum[j])dp[i]=max(dp[i],dp[j]+1);
                else if(num[i]==num[j])dp[i]=max(dp[i],dp[j]);
            }
            if(res