Project 4:Longest Ordered Subsequence

Problem 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 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 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

完成程序

#include <stdio.h>
#define N 10000
int main()
{
    int i,n,ans=1,j;
    int num[N];
    int a[N];
    int max;
    scanf("%d", &n);

    for(i=0;i<n;i++)
    {
        max=0;
        scanf("%d", &num[i]);
        for(j=0;j<i;j++)
        {
            if(num[j]<num[i] && a[j]>max)
                max=a[j];
        }
        a[i]=max+1;
        if(a[i]>ans) ans=a[i];
    }
    printf("%d\n", ans);

    return 0;
}

使用动态规划后运行时间大有下降

首先我们定义变量数组a[N]来保存第n个数为最大数的最长上升子列。动态规划的思路为每增加一个数就判断以该数为最大数最长上升子列，然后与最长的ans做比较。如何找到该数的最长上升子列呢？首先，这个数要比该子列的前面一个数大即为num[j]小于num[i]，这样才能成为最长上升子列的最后一个数。然后,该子列还需是在所有符合第一个条件的子列中最长的一个。及为变量max所起的作用。至此动态规划完毕。

通过全部子列列出后对比长度的解法

#include <stdio.h>
int main()
{
    int N,i,tmp,j,num,max=1,k;
    int a[1000];
    scanf("%d",&N);
    {
        for(i=0;i<N;i++)
            scanf("%d",&a[i]);
        for(i=0;i<N;i++)
        {
            tmp=a[i];
            for(k=1;k<N-i;k++)
            {
                num=1;
                tmp=a[i];
                for(j=i+k;j<N;j++)
                {
                    if(a[j]>tmp)
                        tmp=a[j],num++;
                }
                if(num>max) max=num;
            }
        }
        printf("%d\n",max);
    }
    return 0;
}

这个效率会低很多。