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  • POJ 2533 Longest Ordered Subsequence(裸LIS)

    传送门:

    http://poj.org/problem?id=2533

    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 61731   Accepted: 27632

    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
    

    Source

    Northeastern Europe 2002, Far-Eastern Subregion
     
    分析:
    完全裸露的LIS
     
    code:
    #include<stdio.h>
    #define MAXN 1000
    int dp[MAXN+10],a[MAXN+10];//a数组记录输入的序列
    int main()
    {
        int n,i,j;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
           scanf("%d",&a[i]);
        dp[1]=1;
        for(i=2;i<=n;i++)
        {
            int temp=0;
            for(j=1;j<i;j++)
               if(a[i]>a[j])
                  if(temp<dp[j])
                    temp=dp[j];
            dp[i]=temp+1;
        }  
        int maxlen=0;
        for(i=1;i<=n;i++)
            if(maxlen<dp[i])
               maxlen=dp[i];
        printf("%d
    ",maxlen);
        return 0;  
    }
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  • 原文地址:https://www.cnblogs.com/yinbiao/p/9347993.html
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