Greatest Common Increasing Subsequence hdu1423

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3649    Accepted Submission(s): 1147

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

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define clr(x,y) memset(x,y,sizeof(x))
 using namespace std;
 int a[],b[],ans[],an,bn;
 int work()
 {
     memset(ans,,sizeof(ans));
     int i,j,t,maxa,maxan=;
     for(i=; i<=an; i++)
     {
         maxa=;
         for(j=; j<=bn; j++)
         {
             if(a[i]>b[j]&&ans[j]>maxa)maxa=ans[j];
             if(a[i]==b[j]&&ans[j]<maxa+)
                 ans[j]=maxa+;
         }
     }
     for(i=;i<=bn;i++)
         if(ans[i]>maxan)maxan=ans[i];
     return maxan;
 }
 int main()
 {
     int t,i;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&an);
         for(i=; i<=an; i++)scanf("%d",&a[i]);
         scanf("%d",&bn);
         for(i=; i<=bn; i++)scanf("%d",&b[i]);
         printf("%d\n",work());
         if(t)printf("\n");
     }
 }

Source

ACM暑期集训队练习赛（二）