HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_44969196/article/details/97417110

本文探讨了ZOJ2432问题，即寻找两个序列的最长公共递增子序列的长度。通过使用最长上升子序列模板并结合判断两序列元素是否相等的方法，有效地解决了这一问题。

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Greatest Common Increasing Subsequence

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
题目大意是给定两个数字串seq1、seq2，求出它们最长公共递增子序列的长度。
思路：写出最长上升子序列模板，之后只需判断一下两个数是否相等，输出稍微注意一下

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 505
int dp[maxn],a[maxn],b[maxn],n,m;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        scanf("%d",&m);
        for(int i=1; i<=m; i++)
            scanf("%d",&b[i]);
        for(int i=1; i<=n; i++)
        {
            int maxx = 0;
            for(int j=1; j<=m; j++)
            {
                if(a[i] > b[j])
                    maxx = max(maxx, dp[j]);//最长上升子序列模板
                if(a[i] == b[j])//判断是否相等
                    dp[j] = maxx + 1;
            }
        }
        int ans = 0;
        for(int i=1; i<=m; i++)
            ans = max(ans, dp[i]);
        if(t!=0)
            printf("%d\n\n",ans);
        else
            printf("%d\n",ans);
    }
    return 0;
}