LCIS

LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 647    Accepted Submission(s): 302

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

牺牲空间换时间，给了数的范围，并且有着序列必须是ai=a（i-1）+1的规律，所以可以对数dp，分别保存以ai结尾的序列最长的长度；

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int dp1[1000050],dp2[1000050],a[100005],b[100005];
int main()
{
	int t,n,m,i,j,max1;
	scanf("%d",&t);
	memset(dp1,0,sizeof(dp1));
	memset(dp2,0,sizeof(dp2));
	while(t--)
	{
		max1=0;
		scanf("%d %d",&n,&m);
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		for(i=0;i<m;i++)
		scanf("%d",&b[i]);
		for(i=0;i<n;i++)
		{
			if(dp1[a[i]-1])
			dp1[a[i]]=dp1[a[i]-1]+1;
			else
			{
				dp1[a[i]]=1;
			}
			
		}
		for(i=0;i<m;i++)
		{
			if(dp2[b[i]-1])
			dp2[b[i]]=dp2[b[i]-1]+1;
			else
			{
				dp2[b[i]]=1;
			}
		}
		for(i=0;i<n;i++)
		{
			max1=max(max1,min(dp1[a[i]],dp2[a[i]]));//找公共元素，所以只需遍历一遍a[i],如果a[i]这个点不是公共点，min=0，如果是公共点，min则为前面公共元素的个数 
		}
		printf("%d\n",max1);
		for(i=0;i<n;i++)
		dp1[a[i]]=0;
		for(i=0;i<m;i++)
		dp2[b[i]]=0; 
	}
	return 0;
}