HDU 1423

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5384    Accepted Submission(s): 1742

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 <stdio.h>
#include <string.h>
const int MAXN=510;
int a[MAXN],b[MAXN];
int f[MAXN];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        int m;
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
            scanf("%d",&b[i]);
        for(int i=1;i<=n;i++)
        {
            int MAX=0;
            for(int k=1;k<=m;k++)
            {
                if(a[i]>b[k]&&MAX<f[k])
                    MAX=f[k];
                if(a[i]==b[k])
                    f[k]=MAX+1;
            }
        }
        int temp=0;
        for(int i=1;i<=m;i++)
        {
            if(temp<f[i])
                temp=f[i];
        }
        printf("%d\n",temp);
        if(t)printf("\n");
    }
    return 0;
}