CF 500C New Year Book Reading
2015-11-09 18:37
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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
题目大意:一个人要读n本书每本书有个重量wi,要读m天,每天读1本,他每天把要看的书抽出来(把上面的搬开,拿出要读的书再把上面的书放回去),看完以后放到一摞书的最上面,问根据他的阅读顺序怎样初始化书的排列顺序能使他搬书的重量最小,求出这个最小重量
分析:假如拿书的顺序是134321 那么应该初始化为:1 3 4 2 的顺序放书。
付上一个很玄学的代码。
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
题目大意:一个人要读n本书每本书有个重量wi,要读m天,每天读1本,他每天把要看的书抽出来(把上面的搬开,拿出要读的书再把上面的书放回去),看完以后放到一摞书的最上面,问根据他的阅读顺序怎样初始化书的排列顺序能使他搬书的重量最小,求出这个最小重量
分析:假如拿书的顺序是134321 那么应该初始化为:1 3 4 2 的顺序放书。
付上一个很玄学的代码。
#include<iostream>
#include<map>
#include<cstdio>
using namespace std;
int num[1002];
int w[1000];
int main()
{
int n,m;
while(cin >> n >> m)
{
for(int i=0;i<n;i++)
{
cin >> w[i];
}
int ans = 0;
for(int i=0;i<m;i++)
{
scanf("%d",&num[i]);
map<int,int> mp;
for(int j=i-1;j>=0;j--)
{
if(num[j]==num[i])break;
mp[num[j]]++;
if(mp[num[j]]==1)
ans+=w[num[j]-1];
// cout<<num[j]<<" "<<mp[num[j]]<<" "<<ans<<endl;
}
mp.clear();
}
cout<<ans<<endl;
}
return 0;
}
/*
3 5 1 2 3 1 3 2 3 1*/
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