【打CF，学算法——二星级】Codeforces 703B Mishka and trip (统计)

【CF简介】

题目链接：CF 703B

题面：

A - Mishka and trip Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u SubmitStatusPracticeCodeForces 703B

Description

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

17

Input

5 2
3 5 2 2 4
1 4

Output

71

Hint

This image describes first sample case:

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

This image describes second sample case:

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

题意：

    1-n个点，k个点是关键点，关键点会向所有点连边，非关键点只向两边连边，首尾相接，两点之间最多只有一条边，两点间边的代价，是两个点的值的乘积，问所有边的总代价是多少？

解题：

   简单统计所有边的代价是很简单的，但是因为点数较多，这样是会超时的。可以预先统计出所有点的总代价，所有关键点的总代价。计算时，将所有边统计两遍，其中非关键点较好处理，只要向两边连边，而关键点向所有点连两倍边，关键点之间会有四条边，要后期减去，而关键点和不是相邻的非关键点之间就是两倍边，而关键点和相邻的非关键点（如果是的话）是三倍边，要减去一次。最后，统计完之后，将总代价除以2，即为所求。

代码：

#include 
#include 
#define LL long long
#define sz 100005
using namespace std;
//每个点val
LL val[sz];
//是否是capital
bool vis[sz];
//哪些点是capital
int keyp[sz];
int main()
{
	int n,k,tmp;
	//sum1是所有点的和，sum2是capital的和
	LL sum1=0,ans=0,sum2=0,tmp2=0;
	scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
	{
		scanf("%lldn",&val[i]);
		sum1+=val[i];
	}
	for(int i=1;i<=k;i++)
	{
		scanf("%d",&tmp);
		vis[tmp]=1;
		keyp[i]=tmp;
		sum2+=val[tmp];
    }
	//首尾连接
	val[0]=val[n];
	vis[0]=vis[n];
	val[n+1]=val[1];
	vis[n+1]=vis[1];
	//所有边计算两次
    for(int i=1;i<=n;i++)
	{
	  //是capital那么它连出去的边都计算两次
	  if(vis[i])
	  {   
		  ans+=2*(sum1-val[i])*val[i];
		  //如果前一个点不是capital，那么要减去一次，因为非capital点也会向两边连边
	      if(!vis[i-1])
			  ans-=val[i]*val[i-1];
		  if(!vis[i+1])
			  ans-=val[i]*val[i+1];
	  }
	  //非capital点向两边连边
	  else
         ans+=val[i]*(val[i+1]+val[i-1]);
	}
	//关键点因为向所有点都连了两次，关键点之间连了4次，故需减去2次
    for(int i=1;i<=k;i++)
	{
      tmp2+=(sum2-val[keyp[i]])*val[keyp[i]];
	}
	ans-=tmp2;
	//因为所有边都统计了两次
	printf("%lldn",ans/2);
	return 0;
}