codeforces 500c New Year Book Reading 【思维】

 

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.

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.

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.

 

 

题意：有一个人有n本书，计划一年之内要读m次书（一本书可以反复读）。这n本书是叠放在一起的，每次读一本书要把上面的书移开，在把书给取走。并且读完之后要把它放在最上面，要你求出读完所有书的最小移动重量。
思路：不管最开始如何放置，最先读的书一定是在后面读的书的上面，所以只能要求在读某一本书的时候后，保证上面的都是读过的书就可以。所以，这也就确定了最开始的排列顺序——按照读书的顺序。然后再统计每次读的时候所移动书的重量就可以了。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 1e5;
int weight[MAXN], pos[MAXN];
int st[MAXN], num[MAXN];
bool vis[MAXN];
int main() {
    int n, m;
    while (scanf("%d%d", &n, &m) != EOF) {
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= n; i++) {
            scanf("%d", &weight[i]);
        }
        int cnt = 0;
        for (int i = 0; i < m; i++) {
            scanf("%d", &st[i]);
            if (!vis[st[i]]) {
                pos[cnt++] = st[i];
                vis[st[i]] = true;
            }
        }
        int ans = 0;
        memset(vis, 0, sizeof(vis));
        cnt = 0;
        //遍历每天
        for(int i = 0; i < m; i++) {
            int p;
            for (p = 0; pos[p] != st[i]; p++) {
                ans += weight[pos[p]];
            }
            for(int j = p;j >= 1; j--)
                pos[j] = pos[j - 1];
            pos[0] = st[i];
        }
        printf("%d\n", ans);
    }
    return 0;
}

 

 

 

转载于:https://www.cnblogs.com/cniwoq/p/6770790.html