Greatest Common Increasing Subsequence（LICS模板）-优快云博客

本文链接：https://blog.youkuaiyun.com/zhengde152/article/details/81428702

本文介绍了一种解决最长公共递增子序列问题的方法，并提供了一个C++实现示例。该问题要求找到两个数列之间的最长公共递增子序列。

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模板解析：https://blog.youkuaiyun.com/qq_34374664/article/details/77659413

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

题解：给定两个数字串s1、s2，求出它们最长公共递增子序列的长度（可以不是连续的子序列）

代码：

#include <stdio.h>  
#include <string.h>
#include<algorithm>
using namespace std;  
int s1[510], s2[510];  
int l1, l2;  
int dp[510];
int main()
{  
    int t;  
    scanf("%d", &t);  
    while (t--)
	{  
	    int i,j;
        scanf("%d", &l1);  
        for (i = 1; i <= l1; ++i)  
            scanf("%d", &s1[i]);  
        scanf("%d", &l2);  
        for (i = 1; i <= l2; ++i)  
            scanf("%d", &s2[i]);  
		memset(dp, 0, sizeof(dp)); 
		int maxx;
	   for (i = 1; i <= l1; i++)
	   {  
        maxx= 0;  
        for (j = 1; j <= l2; j++)
		{  
            if (s1[i] > s2[j] && maxx< dp[j]) //判断是否为增序，更新最大值 
				maxx= dp[j]; 
            if (s1[i] == s2[j]) //记录增序终止时，增序的长度 
				dp[j] = maxx+ 1;
        }  
    }  
    maxx = 0;  
    for (i = 1; i <= l2; i++)
        if (maxx< dp[i])  
            maxx = dp[i];  
    printf("%d\n",maxx);  
    if (t)  
        printf("\n");  
  }  
}