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  • hdu1423(最长公共递增子序列)

    Greatest Common Increasing Subsequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3365    Accepted Submission(s): 1062


    Problem Description
    This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.
     
    Input
    Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.
     
    Output
    output print L - the length of the greatest common increasing subsequence of both sequences.
     
    Sample Input
    1
     
    5
    1 4 2 5 -12
    4
    -12 1 2 4
     
    Sample Output
    2
     
    #include<iostream>
    #include<string.h>
    using namespace std;
    int a[505],b[505],f[505];
    int main()
    {
        int T,i,n1,n2,j,k,max;
        cin>>T;
        while(T--)
        {
            cin>>n1;
            for(i=1; i<=n1; i++)
                cin>>a[i];
            cin>>n2;
            for(i=1; i<=n2; i++)
                cin>>b[i];
            memset(f,0,sizeof(f));
            for(i=1; i<=n1; i++)
            {
                max=0;
                for(j=1; j<=n2; j++)
                {
                    if(a[i]>b[j]&&max<f[j])
                        max=f[j];
                    if(a[i]==b[j])
                        f[j]=max+1;
                }
            }
            max=0;
            for(i=1; i<=n2; i++)
                if(f[i]>max)
                    max=f[i];
            cout<<max<<endl;
            if(T!=0)
                cout<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lxm940130740/p/3555036.html
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