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;
}