C. New Year Book Reading（CF）模拟

C. New Year Book Reading

time limit per test

memory limit per test

input

output

n books numbered by integers from 1 to n. The weight of the i-th (1 ≤ i ≤ n) book is wi.

n books by stacking them vertically. When he wants to read a certain book x, he follows the steps described below.

1. lifts all the books above book x.
2. x
3. He puts down the lifted books without changing their order.
4. x, he puts book x

m days. In the j-th (1 ≤ j ≤ m) day, he will read the book that is numbered with integer bj (1 ≤ bj ≤ 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.

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

Input

n (2 ≤ n ≤ 500) and m (1 ≤ m ≤ 1000) — the number of books, and the number of days for which Jaehyun would read books.

n space-separated integers w1, w2, ..., wn (1 ≤ wi) — the weight of each book.

m space separated integers b1, b2, ..., bm (1 ≤ bj ≤ n) — the order of books that he would read. Note that he can read the same book more than once.

Output

lift, which can be achieved by rearranging the order of stacked books.

Sample test(s)

input

3 5 1 2 3 1 3 2 3 1

output

12

Note

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

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>

using namespace std;

int p[10010];
int wi[100010];
int n,m;
int read[10010];

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        __int64 cnt = 0;
        stack<int>q1;
        stack<int>q2;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&wi[i]);
        }
        for(int i=1; i<=n; i++)
        {
            p[i] = 1;
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d",&read[i]);
            if(p[read[i]] == 1)
            {
                q1.push(read[i]);
                p[read[i]] = 0;
            }
        }
        int x,y;
        while(!q1.empty())
        {
            x = q1.top();
            q2.push(x);
            q1.pop();
        }
        for(int i=1; i<=m; i++)
        {
            while(!q2.empty())
            {
                x = q2.top();
                if(x != read[i])
                {
                    cnt += wi[x];
                    q1.push(x);
                    q2.pop();
                }
                else
                {
                    y = x;
                    q2.pop();
                    break;
                }
            }
            while(!q1.empty())
            {
                x = q1.top();
                q2.push(x);
                q1.pop();
            }
            q2.push(y);
        }
       printf("%I64d\n",cnt);

    }
    return 0;
}