This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

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 print L - the length of the greatest common increasing subsequence of both sequences.

1

5
1 4 2 5 -12
4
-12 1 2 4

2

 

 

 

#include<stdio.h>
#include<string.h>
int a[510],b[510];
int dp[510];
int main()
{
    int t,m,n,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(i=0;i<m;i++)
            scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        int ans=0,pos;
        for(i=0;i<n;i++)
        {
            pos=0;
            for(j=0;j<m;j++)
            {
            /// 遍历b的同时，找出 j 之前最大的公共子序列所在点 pos，并且要使 b[pos] < a[i]
            /// 这样，当下一句 b[j]==a[i] 时，就可以直接得到 j 点的最大公共子序列了：dp[pos]+1
                if(b[j]<a[i]&&dp[j]+1>dp[pos])  pos=j;
                if(b[j]==a[i]&&dp[j]<dp[pos]+1) dp[j]=dp[pos]+1;
            }
        }
        for(i=0;i<m;i++)
        {
            if(ans<dp[i])
                ans=dp[i];
        }
        printf("%d\n",ans);
        if(t) printf("\n");
    }
    return 0;
}