CodeForces 703B

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Little Mishka is a great traveller and she visited many countries. After thinking about where to travel this time, she chose XXX — beautiful, but little-known northern country.

Here are some interesting facts about XXX:

XXX consists of n cities, k of whose (just imagine!) are capital cities.
All of cities in the country are beautiful, but each is beautiful in its own way. Beauty value of i-th city equals to c_i.
All the cities are consecutively connected by the roads, including 1-st and n-th city, forming a cyclic route 1 — 2 — ... — n — 1. Formally, for every 1 ≤ i < n there is a road between i-th and i + 1-th city, and another one between 1-st and n-th city.
Each capital city is connected with each other city directly by the roads. Formally, if city x is a capital city, then for every 1 ≤ i ≤ n, i ≠ x, there is a road between cities x and i.
There is at most one road between any two cities.
Price of passing a road directly depends on beauty values of cities it connects. Thus if there is a road between cities i and j, price of passing it equals c_i·c_j.

Mishka started to gather her things for a trip, but didn‘t still decide which route to follow and thus she asked you to help her determine summary price of passing each of the roads in XXX. Formally, for every pair of cities a and b (a < b), such that there is a road between a and b you are to find sum of products c_a·c_b. Will you help her?

Input

The first line of the input contains two integers n and k (3 ≤ n ≤ 100 000, 1 ≤ k ≤ n) — the number of cities in XXX and the number of capital cities among them.

The second line of the input contains n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10 000) — beauty values of the cities.

The third line of the input contains k distinct integers id₁, id₂, ..., id_k (1 ≤ id_i ≤ n) — indices of capital cities. Indices are given in ascending order.

Output

Print the only integer — summary price of passing each of the roads in XXX.

Sample Input

Input

4 1
2 3 1 2
3

Output

Input

5 2
3 5 2 2 4
1 4

Output

Hint

This image describes first sample case:

$技术分享$

It is easy to see that summary price is equal to 17.

This image describes second sample case:

$技术分享$

It is easy to see that summary price is equal to 71.

Source

Codeforces Round #365 (Div. 2)

主要是超时的问题；

可以不把所有的遍历一遍；

如果是首都的话把那个点去掉；

首都到个点的距离ａｎｓ＊ｖａｌｕｅ；

#include <cstdio>
#include <map>
#include <algorithm>
#include <string>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#include <list>
#include <cmath>
#include <set>
using namespace std;
const int maxn=100000+10;
typedef long long ll;
int val[maxn];
int vis[maxn];
int main()
{
    memset(vis,0,sizeof(vis));
    int n,k;
    ll sum=0,ans=0;
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&val[i]);
        ans+=val[i];
    }
    for(int i=1; i<n; i++)
    {
        sum+=val[i]*val[i+1];
    }
    sum+=val[n]*val[1];
    int st,ed;
    for(int i=0; i<k; i++)
    {
        int k_;
        scanf("%d",&k_);
        if(k_==1)
            st=n,ed=2;
        else if(k_==n)
            st=n-1,ed=1;
        else
            st=k_-1,ed=k_+1;

        if(!vis[st]&&!vis[ed])//判断点是否被去掉，如果被去掉就不用再减；
        {
            sum+=(ans-val[k_]-val[st]-val[ed])*val[k_];
        }
        else if(!vis[st]&&vis[ed])
        {
            sum+=(ans-val[k_]-val[st])*val[k_];
        }
        else if(vis[st]&&!vis[ed])
        {
            sum+=(ans-val[k_]-val[ed])*val[k_];
        }
        else if(vis[st]&&vis[ed])
        {

            sum+=(ans-val[k_])*val[k_];
        }
        ans-=val[k_];//去掉是首都的点，如果是首都与每个点都有连线；
        vis[k_]=1;//标记；
    }
    printf("%I64d\n",sum);
    return 0;
}