code force 10D dp 最长公共上升子序列

http://vjudge.net/contest/view.action?cid=47807#problem/D

Description

This problem differs from one which was on the online contest.

The sequence a₁, a₂, ..., a_n is called increasing, if a_i < a_i + 1 for i < n.

The sequence s₁, s₂, ..., s_k is called the subsequence of the sequence a₁, a₂, ..., a_n, if there exist such a set of indexes 1 ≤ i₁ < i₂ < ... < i_k ≤ n that a_ij = s_j. In other words, the sequence s can be derived from the sequence a by crossing out some elements.

You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.

Input

The first line contains an integer n (1 ≤ n ≤ 500) — the length of the first sequence. The second line contains n space-separated integers from the range [0, 10⁹] — elements of the first sequence. The third line contains an integer m (1 ≤ m ≤ 500) — the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 10⁹] — elements of the second sequence.

Output

In the first line output k — the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate the elements with a space. If there are several solutions, output any.

Sample Input

Input

7
2 3 1 6 5 4 6
4
1 3 5 6

Output

3
3 5 6 

Input

5
1 2 0 2 1
3
1 0 1

Output

2
0 1 

题目大意：求已知两个数列的最长公共上升子序列。

解题思路：以一个数列为基准进行匹配另一个数列，如果数字相同，就让被匹配的数组对应下标的f[j]=Max1+1，并用p[j]记录其上一个匹配的数字在被匹配串中的位置；如果被匹配的数字比原串对应数字小，且max1比f[j] 要小就记录更新f[j]的值，并用w记录当前下标，如果被匹配的数字比原串对应数字大，不作处理。

源代码：

#include<cstdio>
#include<iostream>
#include <algorithm>
using namespace std;
int n,m,t[5000];
int p[5000];//记录匹配的长度
int f[5000];//记录匹配位置的下标
int a[5000],b[5000],i,w,j,max1;
void outp(int t)//递归输出最长公共上升子序列
{
    if (t!= 0)
        outp(p[t]),printf("%d ",b[t]);
}
int main()
{
    scanf("%d",&n);
    for (i=1; i<=n; i++)
        scanf("%d",&a[i]);
    scanf("%d",&m);
    for (i=1; i<=m; i++)
        scanf("%d",&b[i]);
    for (i=1; i<=n; i++)
    {
        max1=0;
        w=0;
        for (j=1; j<=m; j++)
        {
            if(a[i]>b[j]&&max1<f[j])
                max1=f[j], w=j;
            if(a[i]==b[j])
                f[j]=max1+1,p[j]=w;
        }
    }
    max1=-1;
    for(i=1; i<=m; i++)
        if(f[i]>max1)
        {
            max1=f[i];
            w=i;
        }
    printf("%d\n",max1);
    if (max1!=0)
        outp(w);
    return 0;
}