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！

 1 //增长lcs algorithm

 2 #include<stdio.h>

 3 #include<string.h>

 4 #define maxn 505

 5 int aa[maxn],bb[maxn];

 6 int lcs[maxn];

 7 int main()

 8 {

 9     int test,na,nb,i,j,maxc,res;

10     scanf("%d",&test);

11     while(test--)

12     {

13         scanf("%d",&na);

14         for(i=1;i<=na;i++)

15            scanf("%d",aa+i);

16            scanf("%d",&nb);

17         for(j=1;j<=nb;j++)

18            scanf("%d",bb+j);

19         memset(lcs,0,sizeof(lcs));

20         for(i=1;i<=na;i++)

21         {

22             maxc=0;

23           for(j=1;j<=nb;j++)

24           {

25               if(aa[i]==bb[j]&&lcs[j]<maxc+1)

26                   lcs[j]=maxc+1;

27               if(aa[i]>bb[j]&&maxc<lcs[j])

28                   maxc=lcs[j];

29           }

30         }

31         res=0;

32         for(i=1;i<=nb;i++)

33            if(res<lcs[i])res=lcs[i];

34            printf("%d\n",res);

35            if(test) putchar(10);

36     }

37     return 0;

38 }