POJ-2533-Longest Ordered Subsequence(LIS模板）

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

注意：一般memset对数组赋0或-1，赋值其他的要用循环来实现

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    int n,i,j,a[1005],d[1005];
    while(scanf("%d",&n)!=EOF)
    {
        int sum=0;

        for(i=0; i<=n; i++)
        {
            d[i]=1;
        }

        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            for(j=1; j<i; j++)
            {
                if(a[j]<a[i])
                    d[i]=max(d[i],d[j]+1);//依次遍历在它以前的元素，找出各个元素中标记的最长长度
            }
            sum=max(sum,d[i]);
        }
        printf("%d\n",sum);
    }
    return 0;
}