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

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Gxjun 2022-03-18 13:56:58 博主文章分类：数据结构 ©著作权

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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 }