自定义序列顺序思想 Permutations

Permutations

Time Limit: 2000ms

Memory Limit: 262144KB

64-bit integer IO format:  %I64d      Java class name:  (Any)

Submit  Status  PID: 20924

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.

Sample Input

Input

3
3 2 1
1 2 3

Output

2

Input

5
1 2 3 4 5
1 5 2 3 4

Output

1

Input

5
1 5 2 3 4
1 2 3 4 5

Output

3

借鉴别人的链接:http://www.cnblogs.com/kkrisen/p/3195673.html

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！

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a[200100],b[200100],aim[200100];
int main()
{
    int n,k;
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++){
           scanf("%d",&k);
           aim[k]=i;
        }
        for(int i=1;i<=n;i++)           b[i]=aim[a[i]];
        int j;
        for (j=2;j<=n;j++){
            if (b[j-1] > b[j])  break;
        }
        printf("%d\n",n-j+1);
    }
    return 0;
}

自己的

#include<stdio.h> 
#include<string.h>
int a[1200000],b[1200000]; 
int main()
{
	int n;
	while(~scanf("%d",&n)){
		for(int i=0;i<n;i++)	scanf("%d",&a[i]);
		for(int i=0;i<n;i++)	scanf("%d",&b[i]);
		int ans=0;
		int p=0;
		for(int i=0;i<n;i++){
			while(a[i]!=b[p] && p<n)
				p++;
			if(a[i]==b[p]){
				ans++;
				p++;
			}
			if(p==n)	break;
		}
		printf("%d\n",n-ans);
	}
	return 0;
}