poj-2533-Longest Ordered Subsequence

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：找出数组中的最长递增序列
简单DP，LIS

代码：

#include <stdio.h>
#include <string.h>
#define max(a,b) a>b?a:b
int main()
{
    int n;
    int i,j;
    while(~scanf("%d",&n))
    {   
        int a[1010],dp[1010];
                //dp[i]表示第0到第i个数字递增序列最大值 

        for(i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        dp[0]=1;

        int max=0;
        for(i=0;i<n;i++){
            dp[i]=1;//初始化dp值都为0 
            for(j=0;j<i;j++){
                if(a[j]<a[i]){
                    dp[i]=max(dp[i],dp[j]+1);
                }
            }
            if(max<dp[i]){
                max=dp[i];
            }
        }
        printf("%d\n",max);
    }
    return 0;
}