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  • poj2533--Longest Ordered Subsequence(dp:最长上升子序列)

    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 33943   Accepted: 14871

    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
    求最长上升子序列:
    dp的求法,初始化时能够将a[0]初始化成一个比全部数小的值,或者是将dp[]全清为1,由于最长上升子序列中,会包括自身,所以最小为1
     
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int a[12000] , dp[12000] ;
    int main()
    {
        int i , j , n , max1 ;
        while(scanf("%d", &n)!=EOF)
        {
            memset(dp,0,sizeof(dp));
            a[0] = -1 ;
            for(i = 1 ; i <= n ; i++)
                scanf("%d", &a[i]);
            for(i = 1 ; i <= n ; i++)
                for(j = 0 ; j < i ; j++)
                    if( a[j] < a[i] && dp[j]+1 > dp[i] )
                        dp[i] = dp[j] + 1 ;
            max1 = 0 ;
            for(i = 1 ; i <= n ; i++)
                max1 = max(max1,dp[i]);
            printf("%d
    ", max1);
        }
        return 0;
    }
    

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  • 原文地址:https://www.cnblogs.com/blfshiye/p/4286986.html
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