LCS

LCS

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 305    Accepted Submission(s): 146

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

一个环的问题，根据题意可知，出现环answer += len() - 1;没有环，则answer += len();

并且若没有环即上下数字相同，即len() = 1;

简单的搜索一下环就可以了。

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn = 1e5+5;
using namespace std;
int a[maxn];
int b[maxn];
int sum[maxn];
bool vis[maxn];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i]);
        }
        for(int i = 1; i <= n; i++){
            scanf("%d",&b[a[i]]);
        }
        for(int i = 1; i <= n; i++){
            vis[i] = 0;
        }
        int ans = 0;
        for(int i = 1; i <= n; i++){
            int x = a[i];
            if(vis[x])continue;
            int cnt = 0;
            while(!vis[x]){
                vis[x] = 1;
                x = b[x];
                cnt++;
            }
            if(cnt == 1)ans++;
            else ans += cnt-1;
        }
        printf("%d
",ans);
    }
    return 0;
}