首页 > 其他 > 详细

POJ-2533-Longest Ordered Subsequence(LIS模板)

时间:2015-07-16 19:04:08      阅读:150      评论:0      收藏:0      [点我收藏+]

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

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;
}

 

POJ-2533-Longest Ordered Subsequence(LIS模板)

原文:http://www.cnblogs.com/tianmin123/p/4651836.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!