hdu 5495 LCS 置换群

LCS

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 207    Accepted Submission(s): 105

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

 

Source

BestCoder Round #58 (div.2)

有几个循环置换，且长度大于2，那么n-多少就是答案

#include
#include
#include
#include
using namespace std;

#define maxn 100007
int a[maxn],b[maxn],vis[maxn];
int dfs(int x){
    vis[x] = 1;
    if(vis[b[a[x]]]) return 1;
    return dfs(b[a[x]])+1;
}
int main(){
    int t,n,u;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 1;i <= n; i++){
            scanf("%d",&u);
            a[u] = i;
        }
        for(int i = 1;i <= n; i++)
            scanf("%d",&b[i]);
        memset(vis,0,sizeof(vis));
        int ans = n;
        for(int i = 1;i <= n; i++){
            if(vis[i] == 0) {
                if(dfs(i) > 1) ans--;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}