Permutations CodeForces - 189C（思维）

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本文介绍了一个有趣的算法问题，挑战者需要将一个数字排列通过特定的操作转换成另一个排列，并找出所需的最小操作次数。文章详细解析了问题背景及解决策略，包括一种创新的方法——通过自定义排序规则简化问题复杂度。

Permutations CodeForces - 189C

Happy PMP is freshman and he is learning about algorithmic problems. He enjoys playing algorithmic games a lot.

One of the seniors gave Happy PMP a nice game. He is given two permutations of numbers 1 through n and is asked to convert the first one to the second. In one move he can remove the last number from the permutation of numbers and inserts it back in an arbitrary position. He can either insert last number between any two consecutive numbers, or he can place it at the beginning of the permutation.

Happy PMP has an algorithm that solves the problem. But it is not fast enough. He wants to know the minimum number of moves to convert the first permutation to the second.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·10⁵) — the quantity of the numbers in the both given permutations.

Next line contains n space-separated integers — the first permutation. Each number between 1 to n will appear in the permutation exactly once.

Next line describe the second permutation in the same format.

Output

Print a single integer denoting the minimum number of moves required to convert the first permutation to the second.

Example

Input

3
3 2 1
1 2 3

Output

Input

5
1 2 3 4 5
1 5 2 3 4

Output

Input

5
1 5 2 3 4
1 2 3 4 5

Output

Note

In the first sample, he removes number 1 from end of the list and places it at the beginning. After that he takes number 2 and places it between 1 and 3.

In the second sample, he removes number 5 and inserts it after 1.

In the third sample, the sequence of changes are like this:

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

So he needs three moves.

大神思路：

给目标序列强制排序，也就是说，不管目标序列里的数字有多乱，我反正按数组下标，强制把目标序列定义为有序数列，比如说1 5 2  3 4，我就反正按数组下标，1就对应第一，5就对应第二。。。然后回到原序列，1 2 3 4 5，根据我们自己定义的排序规则，即按照目标序列排序法，它的排序后的数列应该是 1 3 4 5 2.。。而且这个题目故意说每次只能挑选最后一位数字对序列任意一位进行插入，其实是降低了难度，因为我规定了目标序列是有序数列，其他数列就都得按照目标数列为标准看齐，所以，题目的要求这时候就变成 只要原序列转化出来的新数列是按升序排列，即可，也就是说，求出新数列的最长上升子序列，。。剩下的数字肯定需要通过插入到前方有序数列。。。因此，只要求出最长上升子序列的长度，剩下了多少数字，那就需要插几次。。。原题立解！

简单的总结这种思想，就是自定义规则，我不管目标序列有多乱，我就以他为标准，我就认可他是上升的序列，然后其他序列都得按照他的标准来排序。。。这样就行了

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 200005
int source[maxn];
int aim[maxn];
int temp[maxn];

int main(){
    int n,node;
    scanf("%d",&n);
    int i;
    for(i = 1; i <= n; i++)
        scanf("%d",&source[i]);
    for(i = 1; i <= n; i++){
        scanf("%d",&node);
        aim[node] = i;
    }
    for(i = 1; i <= n; i++){
        temp[i] = aim[source[i]];
    }
    for(i = 2; i <= n; i++){
        if(temp[i-1] > temp[i])
            break;
    }
    printf("%d\n",n-i+1);
    return 0;

}