Santa has to send presents to the kids. He has a large stack of n presents, numbered from 1 to n; the topmost present has number a1, the next present is a2, and so on; the bottom present has number an. All numbers are distinct.
Santa has a list of m distinct presents he has to send: b1, b2, …, bm. He will send them in the order they appear in the list.
To send a present, Santa has to find it in the stack by removing all presents above it, taking this present and returning all removed presents on top of the stack. So, if there are k presents above the present Santa wants to send, it takes him 2k+1 seconds to do it. Fortunately, Santa can speed the whole process up — when he returns the presents to the stack, he may reorder them as he wishes (only those which were above the present he wanted to take; the presents below cannot be affected in any way).
What is the minimum time required to send all of the presents, provided that Santa knows the whole list of presents he has to send and reorders the presents optimally? Santa cannot change the order of presents or interact with the stack of presents in any other way.
Your program has to answer t different test cases.
Input
The first line contains one integer t (1≤t≤100) — the number of test cases.
Then the test cases follow, each represented by three lines.
The first line contains two integers n and m (1≤m≤n≤105) — the number of presents in the stack and the number of presents Santa wants to send, respectively.
The second line contains n integers a1, a2, …, an (1≤ai≤n, all ai are unique) — the order of presents in the stack.
The third line contains m integers b1, b2, …, bm (1≤bi≤n, all bi are unique) — the ordered list of presents Santa has to send.
The sum of n over all test cases does not exceed 105.
Output
For each test case print one integer — the minimum number of seconds which Santa has to spend sending presents, if he reorders the presents optimally each time he returns them into the stack.
Example
inputCopy
2
3 3
3 1 2
3 2 1
7 2
2 1 7 3 4 5 6
3 1
outputCopy
5
8
题意:给你n个物品,每个物品都有序号，放在堆栈中。你有一个清单，清单上有m个物品必须顺序
取出(必须在堆栈中把这件物品以上的都取出来，再把这件物品取出，然后剩下的放回去)一共有
2*(k-1)+1次。但是你可以给取出来的那一大部分排个序，使花费最少。
解析:用一个mx记录向下拿到的最深位置,位于该位置之上的，对答案的贡献只有1.
位于之下的对答案贡献为 2*(该位置-i)+1
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
int a[N],b[N],pos[N];
int f[N];
int t,n,m,x;
typedef long long ll;
map<int,int> v;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
pos[a[i]]=i;
}
int mx=0; //记录深度
ll ans=0;
for(int i=1;i<=m;i++)
{
cin>>x;
if(pos[x]>mx)
{
ans+=(pos[x]-i)*2+1;
}
else
{
ans++;
}
mx=max(mx,pos[x]);
}
cout<<ans<<endl;
}
}
