POJ 2533 : Longest Ordered Subsequence

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Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30983		Accepted: 13518

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_{i_K}), 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

这题和拦截导弹一样，dp求最长递增子序列长度。

wa了好几次，原因是sum(最长长度)初始化错了，开始初始化为0，结果一直wa,后来发现只有一个数时，sum没有经过递归比较直接输出0，但答案应该为1，所以sum的初始化应该应注意。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define max(a,b) a>b?a:b;
int dp[10005];
int a[10005];
int main()
{
    int n;
    scanf("%d",&n);
    int sum =  0;
    for(int i = 0 ; i < n ; i++)
    {
        scanf("%d",&a[i]);
        dp[i]=1;
    }
    for(int i = 1 ; i < n ; i++){
        for(int j = 0 ; j < i ; j++)
        {
            if(a[i]>a[j] && dp[i] < dp[j]+1)
                dp[i] = dp[j]+1;
        }
        sum=max(sum,dp[i]);
    }
//    for(int i = 0 ; i < n  ;i++)
//    printf("%d ",dp[i]);
//    puts("");
    printf("%d\n",sum);
    return 0;
}