CF 500 C. New Year Book Reading 貪心 簡單題

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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.

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 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.

Sample test(s)input

3 5
1 2 3
1 3 2 3 1

output

12

 

 

 

 

題意：有n本書，編號為1~n，每一本書的重量為w[i]，這堆書豎直堆在一起

現在有m天，第i天會從中拿出編號為day[i]的書看，代價是day[i]上面的書的總重量，書一天看完，看完後放在最上面。

 

問：最開始的時候你按照什麼順序放書，會讓你這m天要搬的書的總重量最小。

 

貪心可知，拿每一本書第一次出現的序列作為初始序列，沒有出現的序列放在最下面，不用管了。

 

用一個vector存放這個序列，看完後放在最上面刪除，插入到最上面。

 

 

 1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 #include<iostream> 5  6 using namespace std; 7  8 #define ll long long 9 #define pb push_back10 11 bool vis[505];12 int w[505];13 int day[1005];14 vector<int>v;15 16 int main()17 {18     int n,m;19     scanf("%d %d",&n,&m);20     for(int i=1;i<=n;i++){21         scanf("%d",&w[i]);22     }23     memset(vis,false,sizeof vis);24     for(int i=1;i<=m;i++){25         scanf("%d",&day[i]);26         if(!vis[day[i]]){27             v.pb(day[i]);28             vis[day[i]]=true;29         }30     }31     int ans=0;32     for(int i=2;i<=m;i++){33         int j=0;34         for(;v[j]!=day[i];j++){35             ans+=(ll)(w[v[j]]);36         }37         v.erase(v.begin()+j);38         v.insert(v.begin(),day[i]);39     }40 41     printf("%d\n",ans);42 43     return 0;44 }

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CF 500 C. New Year Book Reading 貪心 簡單題