CodeForces - 500C

最新推荐文章于 2018-08-12 21:58:16 发布

原创最新推荐文章于 2018-08-12 21:58:16 发布 · 451 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #算法 #函数 #贪心 #c语言

贪心算法专栏收录该内容

18 篇文章

订阅专栏

本文介绍了一个有趣的问题：如何通过调整书籍堆叠顺序，最小化在新年阅读计划中所需的体力劳动。通过对阅读序列的理解和合理安排，可以显著减少拿起和放下书籍的总重量。

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 wi.

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.

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

$\text{[math]}$

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 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.

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, ..., wn (1 ≤ wi ≤ 100) — the weight of each book.

The third line contains 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

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

Examples

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.

$\text{[math]}$

解题思路：每看过一本书都会把书放在最上面，所以不管重复看某一本书多少遍，看过一次后它的位置就确定了

我们能做的就是把看一次之前的顺序排好，使其最优。

那就是按照看的顺序来放。

代码：

#include<stdio.h>
#include<string.h>

int wi[505];
int vi[505];
int order[505];
int read[1005];
int main()
{
	int n,m;
	int sum;
	while(~scanf("%d%d",&n,&m))
	{
		memset(vi,0,sizeof(vi));
		sum=0;
		for(int i=0;i<n;i++)
			scanf("%d",&wi[i]);
		int index=0;
		for(int i=0;i<m;i++)
		{
			scanf("%d",&read[i]);
			if(!vi[read[i]])
			{
				order[index++]=read[i];
				vi[read[i]]=1;
			}
		}
		for(int i=0;i<m;i++)
		{
			int j=0;
			for(;read[i]!=order[j];j++)
			{
				sum+=wi[order[j]-1];
			}
			for(int k=j;k>0;k--)
				order[k]=order[k-1];
			order[0]=read[i];
		}
		printf("%d\n",sum);
	}
	
	return 0;
}