hdu 5495 置换

LCS

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 927    Accepted Submission(s): 520

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 integerT, 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 n integers 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

231 2 33 2 161 5 3 2 6 43 6 2 4 5 1

 

Sample Output

24

题意：

给出两个1～n到全排列，问通过同一种置换，得到的两个序列的最长公共子序列最长

题解：

先在草稿上运算两个样例

因为两个序列要经过同一种置换，所以我们可以对第二个序列按照第一个序列进行置换

然后写出这个序列的循环节，发现ans等于所有循环节长度 L   的max（1，L-1）的和

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 100005int a[maxn],b[maxn],vis[maxn];int main(){int x,n,T;freopen("in.txt","r",stdin);scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]);for(int i=1;i<=n;i++) scanf("%d",&x),b[a[i]]=x;int ans=0;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){if(vis[i]) continue;int p=i,cnt=0;do{p=b[p];vis[p]=1;cnt++;}while(p!=i);ans+=max(1,cnt-1);}printf("%d\n",ans);}return 0;}