CF 500 C New Year Book Reading

New Year Book Reading

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

New Year is coming, and Jaehyun decided to read many books during 2015, unlike this year. He has n books numbered by integers from 1 to n. The weight of the i-th (1 ≤ i ≤ n) book is w_i.

As Jaehyun‘s house is not large enough to have a bookshelf, he keeps the n books by stacking them vertically. When he wants to read a certain book x, he follows the steps described below.

1. He lifts all the books above book x.
2. He pushes book x out of the stack.
3. He puts down the lifted books without changing their order.
4. After reading book x, he puts book x on the top of the stack.

He decided to read books for m days. In the j-th (1 ≤ j ≤ m) day, he will read the book that is numbered with integer b_j (1 ≤ b_j ≤ n). To read the book, he has to use the process described in the paragraph above. It is possible that he decides to re-read the same book several times.

After making this plan, he realized that the total weight of books he should lift during m days would be too heavy. So, he decided to change the order of the stacked books before the New Year comes, and minimize the total weight. You may assume that books can be stacked in any possible order. Note that book that he is going to read on certain step isn‘t considered as lifted on that step. Can you help him?

Input

The first line contains two space-separated integers n (2 ≤ n ≤ 500) and m (1 ≤ m ≤ 1000) — the number of books, and the number of days for which Jaehyun would read books.

The second line contains n space-separated integers w₁, w₂, ..., w_n (1 ≤ w_i ≤ 100) — the weight of each book.

The third line contains m space separated integers b₁, b₂, ..., b_m (1 ≤ b_j ≤ n) — the order of books that he would read. Note that he can read the same book more than once.

Output

Print the minimum total weight of books he should lift, which can be achieved by rearranging the order of stacked books.

Sample Input

Input

3 51 2 31 3 2 3 1

Output

12

Hint

Here‘s a picture depicting the example. Each vertical column presents the stacked books.

【思路】看下面copy的不懂大神的思路

题目分析：yy出奇迹，就拿每本书第一次出现的序列作为最终的序列然后按顺序模拟一下，
后来仔细思考了一下理解了这样做的正确性，其实也挺好理解的，
先看一个序列1 1 1 1 1 1 1 2，对于这个序列，我们很明显的策略是按1 2来排，
因为看过以后放在最上面，所以每次读1都不用搬书，对于一个阅读序列的子序列a b，
我们有两种排法，按a b排代价是w[b]，按b a排代价是w[a] + w[b]，
而且要明确的一点是确定了顺序，代价也就随之确定了，我们再将这个性质推广到一般得出贪心结论：
拿每本书第一次出现的顺序作为初始序列，比如1 1 3 1 1 3 2 1 1
按此策略得到的初始序列为1 3 2，不管初始序列是什么样的，
读完 1 1 3以后的书的顺序肯定都是确定的都为3 1 2，但是只有初始序列为1 3 2时得到3 1 2的代价最小，
其实还是要自己理解理解～

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 int w[505], b[1005];
 5 bool vis[505];
 6 int main()
 7 {
 8     int n, m, ans = 0;
 9     scanf("%d %d", &n, &m);
10     for(int i = 1; i <= n; i++)
11         scanf("%d", &w[i]);//the weight of each book.
12     for(int i = 1; i <= m; i++)
13         scanf("%d", &b[i]);// the order of books that he would read
14     for(int i = 2; i <= m; i++)
15     {
16         memset(vis, false, sizeof(vis));
17         for(int j = i - 1; j > 0; j--)
18         {
19             if(b[j] == b[i]) //只要找同一本书读到两次之间的书
20                 break;
21             //对于其上面的书，我只关心它们的总重量，比如1 2 3 2 3 1
22             //我只关心第二次读到1时它上面有哪些书和它们的重量即w[2] + w[3]
23             //这里用个vis标记一下重复的即可
24             if(!vis[b[j]])
25             {
26                 ans += w[b[j]];
27                 vis[b[j]] = true;
28             }
29         }
30     }
31     printf("%d\n", ans);
32 }