LCIS

 LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 499    Accepted Submission(s): 229

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
  

 

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <cstring>
using namespace std;

typedef long long ll;

const int maxn = 1000005;
int a[maxn],b[maxn];

int main(){
	
	int T;
	scanf("%d",&T);
	int num;
	int N,M;
	int Max;
	while(T--){
		Max = 0;
		scanf("%d %d",&N,&M);
		for(int i=1; i<=N; i++){
			scanf("%d",&num);
			Max = max(Max,num);
			a[num] = a[num-1] + 1;
		}
		for(int i=1; i<=M; i++){
			scanf("%d",&num);
			Max = max(Max,num);
			b[num] = b[num-1] + 1;
		}
		int ans = 0;
		for(int i=1; i<=Max; i++){
			ans = max(ans,min(a[i],b[i]));
			a[i] = b[i] = 0;
		}
		printf("%d\n",ans);
	}
	return 0;
}