【codeforces】New Year Book Reading

川长思鸟来 2022-09-26 12:57 176阅读 0赞

New Year Book Reading

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

Submit [Status][]

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.

![8df1fb9bb3cd7b22d35c2affcdccf2a0][]

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 *w*1, *w*2, ..., *w**n* (1 ≤ *w**i* ≤ 100) — the weight of each book.

The third line contains *m* space separated integers *b*1, *b*2, ..., *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 5
    1 2 3
    1 3 2 3 1

Output

Hint

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

![64c2c8479b6fdf741903770c3296cf70][]

解决本题用栈的结构最容易理解，开两个栈配合使用。关键就是按他给的读书的顺序（这样就能保证所举起的总重量最小）使书的序号入第一个栈，接下来操作就很简单了，代码一看便知。

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<stack>
    #include<algorithm>
    using namespace std;
    const int N = 505;
    int w[N],d[N*2];
    bool vis[N];
    int main() {
    	int n,m;
    	while(scanf("%d%d",&n,&m)!=EOF) {
    		stack<int >u;
    		stack<int >v;
    		memset(vis,0,sizeof(vis));
    		for(int i=1; i<=n; i++) {
    			scanf("%d",&w[i]);
    		}
    		int k;
    		for(int i=1; i<=m; i++) {
    			scanf("%d",&d[i]);
    			if(!vis[d[i]]) {
    				u.push(d[i]);
    				vis[d[i]]=1;
    			}
    		}
    		while(!u.empty()) {
    			int x=u.top();
    			u.pop();
    			v.push(x);
    		}
    		int ans=0;
    		for(int i=1; i<=m; i++) {
    			int t=d[i];
    			while(v.top()!=t) {
    //				if(v.top()==t) {
    //					break;
    //				}
    				int x=v.top();
    				v.pop();
    				ans+=w[x];
    				u.push(x);
    			}
    			v.pop();
    			while(!u.empty()) {
    				int x=u.top();
    				u.pop();
    				v.push(x);
    			}
    			v.push(t);
    		}
    		printf("%d\n",ans);
    	}
    	return 0;
    }

[Status]: http://vjudge.net/contest/132750#status//C/0
[8df1fb9bb3cd7b22d35c2affcdccf2a0]: http://7xjob4.com1.z0.glb.clouddn.com/8df1fb9bb3cd7b22d35c2affcdccf2a0
[64c2c8479b6fdf741903770c3296cf70]: http://7xjob4.com1.z0.glb.clouddn.com/64c2c8479b6fdf741903770c3296cf70