$O(n)$ 差分

用差分算出每个房子经过的次数，计算出每个房子路过的花费，比较购买花费和路过花费选择更小的加入答案。

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;

long long n, p[N], a[N], b[N], d[N];

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> p[i];
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        cin >> b[i];

    for (int i = 1; i <= n; i++)
    {
        int l = min(p[i - 1], p[i]), r = max(p[i - 1], p[i]);
        if (p[i - 1] == l)
        {
            d[l + 1]++;
            d[r + 1]--;
        }
        else
        {
            d[l]++;
            d[r]--;
        }
    }
    d[p[n] + 1]++;
    d[n + 1]--;

    long long res = 0;
    for (int i = 1; i <= n; i++)
    {
        d[i] += d[i - 1];
        res += min(a[i], b[i] * d[i]);
    }
    cout << res;

    return 0;
}