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=1;i<=na;i++)
           scanf("%d",aa+i);
           scanf("%d",&nb);
        for(j=1;j<=nb;j++)
           scanf("%d",bb+j);
        memset(lcs,0,sizeof(lcs));
        for(i=1;i<=na;i++)
        {
            maxc=0;
          for(j=1;j<=nb;j++)
          {
              if(aa[i]==bb[j]&&lcs[j]<maxc+1)
                  lcs[j]=maxc+1;
              if(aa[i]>bb[j]&&maxc<lcs[j])
                  maxc=lcs[j];
          }
        }
        res=0;
        for(i=1;i<=nb;i++)
           if(res<lcs[i])res=lcs[i];
           printf("%d
",res);
           if(test) putchar(10);
    }
    return 0;
}