题目链接：

LCS

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 818    Accepted Submission(s): 453

Problem Description

You are given two sequence {a1,a2,...,an} and {b1,b2,...,bn}. Both sequences are permutation of {1,2,...,n}. You are going to find another permutation {p1,p2,...,pn} such that the length of LCS (longest common subsequence) of {ap1,ap2,...,apn} and {bp1,bp2,...,bpn} is maximum.

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 an integer n(1≤n≤105) - the length of the permutation. The second line contains n integers a1,a2,...,an. The third line contains nintegers b1,b2,...,bn.

The sum of n in the test cases will not exceed 2×106.

Output

For each test case, output the maximum length of LCS.

Sample Input

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

Sample Output

2
4

题意：

问把数列重新排一下的LCS的长度是多少;

思路：

可以发现把置换分成循环后除长度为一的循环外,每个循环都可以变换最后形成l-1的LCS,所以就好了;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,a[maxn],b[maxn],pos[maxn],vis[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]),pos[a[i]]=i,vis[i]=0;
        for(int i=1;i<=n;i++)scanf("%d",&b[i]);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(!vis[b[i]])
            {
                vis[b[i]]=1;
                int len=0,fa=b[i],p;
                while(1)
                {
                    p=pos[fa];
                    fa=b[p];
                    len++;
                    if(vis[fa])break;
                    vis[fa]=1;
                }
                if(len==1)ans++;
                else ans+=len-1;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}