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1000 ms  |  内存限制： 65535

3

There are A, B two sequences, the number of elements in the sequence is n、m;

Each element in the sequence are different and less than 100000.

Calculate the length of the longest common subsequence of A and B.

The input has multicases.Each test case consists of three lines;

The first line consist two integers n, m (1 < = n, m < = 100000);

The second line with n integers, expressed sequence A;

The third line with m integers, expressed sequence B;

输出 For each set of test cases, output the length of the longest common subsequence of A and B, in a single line. 样例输入

5 4 1 2 6 5 4 1 3 5 4

样例输出

3

//函数lower_bound()在first和last中的前闭后开区间进行二分查找， //返回大于或等于val的第一个元素位置。如果所有元素都小于val， //则返回last的位置，且last的位置是越界的！返回查找元素的第一个 //可安插位置，也就是“元素值>=查找值”的第一个元素的位置

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 100010
using namespace std;
int a[N];
int b[N];
int dp[N];
int main()
{
	int n,m,i,j;
	int x;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&x);
			dp[x]=i;
		}
		int r=0;
		for(i=1;i<=m;i++)
		{
			scanf("%d",&x);
			if(dp[x])
				b[r++]=dp[x];
		}
		int p=0;
		dp[p++]=b[0];
		for(i=1;i<r;i++)
		{
			if(dp[p-1]<b[i])
				dp[p++]=b[i];
			else
			{
				x=lower_bound(dp,dp+p,b[i])-dp;
				dp[x]=b[i];
			}
		}
		printf("%d\n",p);
	}
	return 0;
}