模拟（所有边权）

https://acm.ecnu.edu.cn/contest/231/problem/D/

题意：给你n个点，且给出n个点的点权，给出特殊点（值为1），特殊点有到其他任何边权。

边权等于两点权值乘积。求所以边权之和。

解法：一边遍历一遍筛除特殊点。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include<time.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f63f3f
#define mod 20191117
#define PI acos(-1)
using namespace std;
typedef long long ll ;
ll a[100009];
ll vis[100009];
ll ans = 0, sum = 0 ;


int main()
{
    #ifdef ONLINE_JUDGE
    #else
        freopen("D:/c++/in.txt", "r", stdin);
        freopen("D:/c++/out.txt", "w", stdout);
    #endif
    ll n ;

    scanf("%lld" , &n);
    scanf("%lld" , &a[1]);
    sum = a[1] ;
    for(int i = 2 ; i <= n ; i++)
    {
        scanf("%lld" , &a[i]);
        sum += a[i];
        ans += a[i] * a[i-1];
    }
    ans += a[1] * a[n];
    a[0] = a[n] ;
    a[n+1] = a[1];
    for(int i = 1 ; i <= n ; i++)
    {
        scanf("%lld" , &vis[i]);
    }
    for(int i = 1 ; i <= n - 1 ; i++)
    {
        if(vis[i])
        {
            ll t = a[i-1] + a[i+1];
            if(vis[i-1]) t = a[i+1];
            sum -= a[i] ;
            ans += a[i] * (sum - t);
        }
    }
    if(vis[n])
    {
        if(vis[1]) sum += a[1];//以防筛除两次
        ll t = a[n-1] + a[n+1];
        if(vis[n-1]) t = a[n+1];
        sum -= a[n];
        ans += a[n] * (sum - t);
    }
    cout << ans << endl ;





    return 0;
}