Good Bye 2014 C. New Year Book Reading 贪心
2015-08-22 13:59
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C. New Year Book Reading
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
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
output
Note
Here's a picture depicting the example. Each vertical column presents the stacked books.
确定书的排列,找最小值。
书肯定是按看的顺序出现,这样,需要的代价最小,因为,如果,不是按这样的排列,其代价,会额外增加,把它放到前面的代价。再用list模拟一下就好了,复杂度为o(n * m );
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
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
Note
Here's a picture depicting the example. Each vertical column presents the stacked books.
确定书的排列,找最小值。
书肯定是按看的顺序出现,这样,需要的代价最小,因为,如果,不是按这样的排列,其代价,会额外增加,把它放到前面的代价。再用list模拟一下就好了,复杂度为o(n * m );
#define N 505 #define M 1005 #define maxn 205 #define MOD 1000000000000000007 int n,m,w ,t,b[M]; list<int> l; list<int>::iterator it; bool vis ; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); while(S2(n,m)!=EOF) { fill(vis,false); l.clear(); FI(n) S(w[i]); ll ans = 0; FI(m){ S(t);t--; b[i] = t; if(!vis[t]){ vis[t] = true; l.insert(l.end(),t); } } /* for(it = l.begin();it != l.end();it++){ printf("%d ",*it); } cout<<endl */ FI(m){ t = b[i]; it = l.begin(); if(*it != t){ while(*it != t){ ans += w[*it]; it++; } l.erase(it); l.insert(l.begin(),t); } } cout<<ans<<endl; } //fclose(stdin); //fclose(stdout); return 0; }
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