Greatest Common Increasing Subsequence（最长单调递增公共子序列+hdu1423）

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5557    Accepted Submission(s): 1816

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）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1423  

思路：求最长递增公共子序列模板题

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,m,a[505],b[505];

int LICS()
{
    int ans=0;
    int dp[505]={0};
    for(int i=1;i<=n;i++)
    {
        int len=0;
        for(int j=1;j<=m;j++)
        {
            if(a[i]>b[j])
                len=max(len,dp[j]);
            else if(a[i]==b[j])
                dp[j]=max(dp[j],len+1);

            ans=max(ans,dp[j]);
        }
    }

    return ans;
}

int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        scanf("%d",&m);
        for(i=1;i<=m;i++)
        {
            scanf("%d",&b[i]);
        }
        printf("%d\n",LICS());
        if(t) printf("\n");
    }

    return 0;
}