原題：
C. Stack of Presents
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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≤10^5) — 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 10^5.
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
Input
2
3 3
3 1 2
3 2 1
7 2
2 1 7 3 4 5 6
3 1
Output
5
8
中文：
聖誕老人有一個禮物箱子，還有一個禮物名單。
禮物箱子是一個棧結構，如果禮物在棧頂，拿出一個禮物需要花費1s的時間，否則需要將需要拿出的禮物先一個一個出棧，然後把不需要的禮物放回去，這樣一共需要2*n+1的時間，其中n是在棧中，當前禮物上面的禮物個數。不過，從棧中拿出的禮物，再次入棧時，可以按照任意順序入棧，現在問你按照名單去拿禮物，最少花費多少時間？
代碼：
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<string.h>
#include<unordered_map>
#include<map>
#include<set>
#include<queue>
#include<functional>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int maxn = 100010;
ll a[maxn];
ll b[maxn];
ll pos[maxn];
int t, m, n;
int main()
{
ios::sync_with_stdio(false);
cin >> t;
while (t--)
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
pos[a[i]] = i;
}
for (int i = 1; i <= m; i++)
cin >> b[i];
ll ans = 0,res = 0;
for (int i = 1; i <= m; i++)
{
if (pos[b[i]] <= res)
{
ans++;
}
else
{
ans += (pos[b[i]] - i) * 2 + 1;
res = pos[b[i]];
}
}
cout << ans << endl;
}
return 0;
}
思路：
假設要取出的物品a是名單上第一個禮物，在棧中的第4層。那麼首先把上面3個物品拿出來用3s，然後然後把a拿出來用1秒，把上面3個物品放回去用3s，至於怎麼放進去，不用管。如果名單上的下一個禮物在這3物品當中，直接按找名單中的順序直接出棧即可，默認就是按照名單的順序重新放回棧內。