New Year Book Reading（贪心）

原题链接
C. New Year Book Reading
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

这个题就是要把这些书先开始就按照阅读顺序排号，中间有重复的编号就忽略掉直到把所有编号都存进一个数组里去，再模拟搬书的过程即可

#include <cstdio>#include <algorithm>#include <iostream>#include <vector>using namespace std;int w[510],order[1010],zhuangtai[510];vector<int> a;//a就是那个书堆，其中的某个元素就是某本书的质量在w中所对的编号，而order其实也全是编号int main(){        int n,m;        cin >> n >> m;        for(int i=1;i<=n;i++) cin >> w[i];//因为书的重量的编号就是从1开始的        for(int i=0;i<m;i++){                cin >> order[i];                //order[i]--;//因为对应的weiight                if(zhuangtai[order[i]] == 0){//说明这个数还没有被选过                        a.push_back(order[i]);                        zhuangtai[order[i]]=1;                }        }        long long sum=0;        for(int i=0;i<m;i++){                int p=0;//p就是这个堆上从最上面往下面移动的小标指示                while(a[p] != order[i]){sum += w[a.at(p)];p++;}                vector<int>::iterator it = a.begin() + p;                a.erase(it);//删除目标位置的元素                a.insert(a.begin(),order[i]);//再把书放到最上面去        }        cout << sum <<endl;        return 0;}