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[600],b[600],ans[600],an,bn;
int work()
{
    memset(ans,0,sizeof(ans));
    int i,j,t,maxa,maxan=0;
    for(i=1; i<=an; i++)
    {
        maxa=0;
        for(j=1; j<=bn; j++)
        {
            if(a[i]>b[j]&&ans[j]>maxa)maxa=ans[j];
            if(a[i]==b[j]&&ans[j]<maxa+1)
                ans[j]=maxa+1;
        }
    }
    for(i=1;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=1; i<=an; i++)scanf("%d",&a[i]);
        scanf("%d",&bn);
        for(i=1; i<=bn; i++)scanf("%d",&b[i]);
        printf("%d
",work());
        if(t)printf("
");
    }
}

Source

ACM暑期集训队练习赛（二）