dp（最长公共上升子序列）

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

InputEach sequence is described with M - its length (1 <= M
<= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the
sequence itself.Outputoutput 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

2lis（最长上升子序列）和lcs（最长公共子序列）的结合lcis（最长公共上升子序列）还不是很懂这个问题https://www.cnblogs.com/WArobot/p/7479431.html

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

int w[109] , dp[1009] , a[1009] , b[1009];

int main()
{
    int n ;
    scanf("%d" , &n);
    while(n--)
    {
        int m  , l;
        scanf("%d" , &m);
        memset(dp , 0 , sizeof(dp));
        for(int i = 1 ; i <= m ; i++)
        {
            scanf("%d" , &a[i]);
        }
        scanf("%d" , &l);
        for(int i = 1 ; i <= l ; i++)
        {
            scanf("%d" , &b[i]) ;
        }
        int mas = 0 ;
        for(int i = 1 ; i <= m ; i++)
        {
            mas = 0 ; // 记录b数组前j个与a数组前i个的最长公共升序列的个数
            for(int j = 1 ; j <=l ; j++)
            {
                if(a[i] > b[j])
                    mas = max(mas , dp[j]);
                if(a[i] == b[j])
                    dp[j] = mas + 1 ;
            }
        }
        int ans = 0 ;
        for(int i = 1 ; i <= l ; i++)
        {
            ans = max(ans , dp[i]);
        }
        if(n)
        {
            printf("%d\n\n", ans);
        }
        else
            printf("%d\n" , ans);

    }

    return 0;
}

原文地址：https://www.cnblogs.com/nonames/p/11237948.html