hdu1423Greatest Common Increasing Subsequence（最长公共递增子序列）

Greatest Common Increasing Subsequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 3

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）

题意：

求两个数组的最长公共递增子序列

#include<cstdio>
#include<cstring>
int a[10000],b[10000],dp[1000][1000];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int x1,x2,i,j,max1;
        scanf("%d",&x1);
        for(i=0;i<x1;i++)
        scanf("%d",&a[i]);
        scanf("%d",&x2);
        for(i=0;i<x2;i++)
        scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        for(i=1;i<=x1;i++)
        {
            max1=0;
            for(j=1;j<=x2;j++)
            {
                dp[i][j]=dp[i-1][j];//至少等于前面的
                if(a[i-1]>b[j-1]&&max1<dp[i-1][j])//有可能在公共子序列里面；
                    max1=dp[i-1][j];
                if(a[i-1]==b[j-1])//相等就说明是递增子序列里面的。
                    dp[i][j]=max1+1;
            }
        }
        max1=0;
        for(i=0;i<=x1;i++)//找出较大的
            for(j=0;j<=x2;j++)
            if(max1<dp[i][j])
            max1=dp[i][j];
        printf("%d\n",max1);
        if(n)printf("\n");//注意就是最后的没有换行。
    }


}