# LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 440    Accepted Submission(s): 201

Problem Description
Alex has two sequences a1,a2,…,an and b1,b2,…,bm. He wants find a longest common subsequence that consists of consecutive values in increasing order.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:The first line contains two integers n and m (1≤n,m≤100000) — the length of two sequences. The second line contains n integers: a1,a2,…,an (1≤ai≤106). The third line contains n integers: b1,b2,…,bm (1≤bi≤106).

There are at most 1000 test cases and the sum of n and m does not exceed 2×106.

Output
For each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.

Sample Input
3
3 3
1 2 3
3 2 1
10 5
1 23 2 32 4 3 4 5 6 1
1 2 3 4 5
1 1
2
1

Sample Output
1
5
0

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<iostream>
using namespace std;
#define MMAX 110000
int a[MMAX],b[MMAX],dpa[MMAX],dpb[MMAX];
int main()
{
cout.sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
int n,m;
cin>>n>>m;
memset(dpa,0,sizeof(dpa));
memset(dpb,0,sizeof(dpb));
int maxx=0;
for(int i=1; i<=n; i++)
{
cin>>a[i];
dpa[a[i]]=dpa[a[i]-1]+1;
maxx=max(maxx,a[i]);
}
for(int i=1; i<=m; i++)
{
cin>>b[i];
dpb[b[i]]=dpb[b[i]-1]+1;
maxx=max(maxx,b[i]);
}
int ans=0;
for(int i=1; i<=maxx; i++)
ans=max(ans,min(dpa[i],dpb[i]));
cout<<ans<<endl;
}
}