CODE FORCE 500C (New Year Book Reading) (贪心or 模拟)

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.

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.



题目大意：  n 本书，  每本书都有重量，  m天  一天读一本书   Mj  代表第j天 读 Mj书   读的书 可以重复；

当天要读的书， 要先把 上面的书移走，   问  最少的移动重量是多少；

方法一：  贪心的思想；  把所有的书， 包括天数在内的所有   放在一条直线上，  然后 从2 开始 扫描 知道扫到两个天数一样， 结束，  两者之间的 即为 要移动的重量；

方法二：  模拟操作，  按照读书顺序  放在一摞，  重复的书 只进行第一个操作；  模拟所有的操作；

法一代码：   

#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
#include <stack>
#include <stdio.h>
typedef long long ll;
const int MAXN=10100;
using namespace std;
int a[MAXN],b[MAXN];
int vis[MAXN];
int main()
{
int n,m;
while(cin>>n>>m)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=1;i<=n;i++)
cin>>a[i];
for(int i=1;i<=m;i++)
cin>>b[i];
ll sum=0;
for(int i=2;i<=m;i++)
{
memset(vis,0,sizeof(vis));
for(int j=i-1;j>=1;j--)
{
if(b[j]==b[i])
break;
if(!vis[b[j]])
{
sum += a[b[j]];
vis[b[j]]=1;
}
}
}
cout<<sum<<endl;
}
return 0;
}
/*
5 7
2 4 9 5 6
3 2 4 1 5 2 4
3 5
1 1 1
3 2 1 2 3
*/


法二代码：

#include <iostream>
#include <queue>
#include <cstring>
#include <algorithm>
#include <stack>
#include <stdio.h>
typedef long long ll;
const int MAXN=10100;
struct node{
int we;
int col;
}c[MAXN];

using namespace std;

int a[MAXN],b[MAXN];
int vis[MAXN];
int main()
{
int n,m;
while(cin>>n>>m)
{
memset(vis,0,sizeof(vis));
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=1;i<=n;i++)
cin>>a[i];
for(int i=1;i<=m;i++)
cin>>b[i];

for(int i=1,k=1;i<=m;i++)
{
if(!vis[b[i]]) // 注意重复存储的问题;
{
c[k].we=a[b[i]];// 要用一个新的 k 值存储;
c[k].col=b[i];
vis[b[i]]=1;
k++;
}
}
ll sum=0;
for(int i=1;i<=m;i++)// m天
{
for(int j=1;j<=n;j++)// 所有的书
{
if(c[j].we==a[b[i]]&&c[j].col==b[i])// 结束
{
int temp1=c[j].we;
int temp2=c[j].col;
for(int k=j;k>=1;k--)// 交换
{
c[k].we=c[k-1].we;
c[k].col=c[k-1].col;
}
c[1].we=temp1;
c[1].col=temp2;
break;
}
else
sum+=c[j].we;// 加重量
}
}
cout<<sum<<endl;
}
return 0;
}
/*
5 7
2 4 9 5 6
3 2 4 1 5 2 4
3 5
1 1 1
3 2 1 2 3
*/

123