Huawei written:

Subject description:

Input two arrays of integers A and B, both of the elements satisfy the unique and random, while the elements of A are present in the B, B, is also present in A, i.e., A and B, only the order of elements may be different, For example, (1,3,5,2) and (3,2,1,5).

Are now want to delete part of the element A and B, such that A and B are the same as the rest of the sequence, output a minimum number of elements in the array A to be deleted (note array B need to remove the same number of elements).

Enter a description:

Input common three lines, a first line integer n (1 <= n <= 100000), A and B represents the number of elements. The second line array A, a total of n integers, the behavior of the third array B, a total of n integers.

Output Description:

The output of a number of elements in the array A behavior you want to delete (contain newline).

 

Input:

4

1 3 5 2

3 2 1 5

Output:

2

 

Description:

{1, 3, 5, 2}, and {3, 2, 1, 5} of the three longest common subsequence. Are {1, 5}, {3, 5}, {3, 2}, so the need to remove at least two elements.

 

Analysis of ideas:

First, using a hash table to save each element in the array A, B according to the input of the next, by the A array to hold the position of the corresponding element in the B. For the essence A, B is equal to the need to find elements both increase in the longest sequences.

 

Code:

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<vector>
 4 #include<algorithm>
 5 #include<string>
 6 #include<map>
 7 #include<limits.h>
 8 #include<queue>
 9 using namespace std;
10 
11 int LIS(vector<int>&A, vector<int>&B)
12 {
13     int len=1;
14     B[0]=A[0];
15     for(int i=1; i<A.size(); i++)
16     {
17         if(A[i]>B[len-1])
18             B[len++]=A[i];
19         else
20         {
21             int p = lower_bound(B.begin(), B.begin()+len, A[i])-B.begin();
22             B[p] = A[i];
23         }
24     }
25     return len;
26 }
27 
28 int main()
29 {
30     int n;
31     cin>>n;
32     vector<int>A(n);
33     vector<int>B(n);
34     map<int, int>M;
35     for(int i=0; i<n; i++)
36     {
37         cin>>A[i];
38         M[A[i]]=i;
39     }
40     int k=0;
41     for(int i=0; i<n; i++)
42     {
43         int x;
44         cin>>x;
45         A[M[x]] = k++;
46     }
47     int ans = n-LIS(A,B);
48     cout<<ans<<endl;
49     return 0;
50 }