原题：
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：
圣诞老人有一个礼物箱子，还有一个礼物名单。
礼物箱子是一个栈结构，如果礼物在栈顶，拿出一个礼物需要花费1s的时间，否则需要将需要拿出的礼物先一个一个出栈，然后把不需要的礼物放回去，这样一共需要2*n+1的时间，其中n是在栈中，当前礼物上面的礼物个数。不过，从栈中拿出的礼物，再次入栈时，可以按照任意顺序入栈，现在问你按照名单去拿礼物，最少花费多少时间？
代码：
#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：
假设要取出的物品a是名单上第一个礼物，在栈中的第4层。那么首先把上面3个物品拿出来用3s，然后然后把a拿出来用1秒，把上面3个物品放回去用3s，至于怎么放进去，不用管。如果名单上的下一个礼物在这3物品当中，直接按找名单中的顺序直接出栈即可，默认就是按照名单的顺序重新放回栈内。