HDOJ 题目1423 Greatest Common Increasing Subsequence（LICS）_common subsequence动态规划实验报告-优快云博客

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本文介绍了一个经典的算法问题——最长公共递增子序列（LCIS），并提供了一段AC代码实现。该问题要求找出两个序列中最长的共同递增子序列的长度。输入包含两个整数序列，输出为该子序列的长度。

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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): 4693 Accepted Submission(s): 1496

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

Source

ACM暑期集训队练习赛（二）

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ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int a[550],b[550],n,m,dp[550];
int lics()
{
    int ans=0,i,j;
    for(i=1;i<=n;i++)
    {
        int ma=0;
        for(j=1;j<=m;j++)
        {
            int temp=ma;
            if(b[j]<a[i]&&dp[j]>ma)
                ma=dp[j];
            if(b[j]==a[i])
                dp[j]=temp+1;
            ans=max(ans,dp[j]);
        }
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        //int n,m;
        int i,j;
        scanf("%d",&n);
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(j=1;j<=m;j++)
            scanf("%d",&b[j]);
        printf("%d\n",lics());
        if(t)
            printf("\n");
    }
}