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  • POJ 2533 Longest Ordered Subsequence(DP 最长上升子序列)

    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 38980   Accepted: 17119

    Description

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., 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



    n*n算法

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<queue>
    
    using namespace std;
    
    int n;
    int a[1001];
    int dp[1010];
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            int maxx = -1;
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i=0;i<n;i++)
            {
                dp[i] = 1;
                for(int j=0;j<i;j++)
                {
                    if(a[i]>a[j] && dp[j]+1>dp[i])
                    {
                        dp[i] = dp[j] + 1;
                    }
                }
                if(maxx < dp[i])
                {
                     maxx = dp[i];
                }
            }
            printf("%d
    ",maxx);
        }
        return 0;
    }




    n*logn算法:

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define inf 999999
    
    using namespace std;
    
    int n;
    int dp[1010];
    int a[1010];
    
    int res(int len,int num)
    {
        int l = 0,r = len;
        while(l!=r)
        {
            int mid = (l+r)>>1;
            if(dp[mid] == num)
            {
                return mid;
            }
            else if(dp[mid]<num)
            {
                l = mid + 1;
            }
            else if(dp[mid]>num)
            {
                r = mid;
            }
        }
        return l;
    }
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            int len = 1;
            dp[0] = -1;
            for(int i=1;i<=n;i++)
            {
                dp[i] = inf;
                int k = res(len,a[i]);
                if(k == len)
                {
                    len++;
                }
                dp[k] = a[i];
            }
            printf("%d
    ",len-1);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/zsychanpin/p/6884550.html
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