HDUOJ ---1423 Greatest Common Increasing Subsequence（LCS）

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3460    Accepted Submission(s): 1092

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

Source

ACM暑期集训队练习赛（二）

代码：动态规划求最长增长公共序列 下面展示的是压缩空间的lcs，由于不需要记录顺序，所以这样写，较为简便，如果要记录路径只需要将lcs[]--->换成lcs[][],

然后maxc，变为lcs[][]的上一行即可！

 //增长lcs algorithm
 #include<stdio.h>
 #include<string.h>
 #define maxn 505
 int aa[maxn],bb[maxn];
 int lcs[maxn];
 int main()
 {
     int test,na,nb,i,j,maxc,res;
     scanf("%d",&test);
     while(test--)
     {
         scanf("%d",&na);
         for(i=;i<=na;i++)
            scanf("%d",aa+i);
            scanf("%d",&nb);
         for(j=;j<=nb;j++)
            scanf("%d",bb+j);
         memset(lcs,,sizeof(lcs));
         for(i=;i<=na;i++)
         {
             maxc=;
           for(j=;j<=nb;j++)
           {
               if(aa[i]==bb[j]&&lcs[j]<maxc+)
                   lcs[j]=maxc+;
               if(aa[i]>bb[j]&&maxc<lcs[j])
                   maxc=lcs[j];
           }
         }
         res=;
         for(i=;i<=nb;i++)
            if(res<lcs[i])res=lcs[i];
            printf("%d\n",res);
            if(test) putchar();
     }
     return ;
 }