三分

带两个约束条件，如果你数学好约分化简合并之后就是个抛物线，要求最顶点（拐点）直接三分答案即可，注意check函数的写法，而且C++写的注意爆了ll，可以使用__int128保险一点，我试了下ull也过了。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ull s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ull x) { if (!x) { putchar('0'); return; } char F[40]; ull tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e6 + 7;
ull a, b, c;
ull n, m;
ull t[N], g[N];

ull check(ull x) {
    ll p = 0, q = 0, res = 0;
    for (int i = 1; i <= m; i++)
        if (g[i] >= x)    p += g[i] - x;
        else q += x - g[i];
    if (b <= a)res = p * b;
    else {
        if (q >= p)    res = p * a;
        else res = q * a + (p - q) * b;
    }
    for (int i = 1; i <= n; i++)
        if (t[i] < x)    res += (x - t[i]) * c;
    return res;
}

int main() {
    a = read(), b = read(), c = read();
    n = read(), m = read();
    for (int i = 1; i <= n; i++)    t[i] = read();
    for (int i = 1; i <= m; i++)    g[i] = read();
    ull l = 1, r = *max_element(g + 1, g + m + 1);
    while (l + 10 < r) {
        ull m = l + r >> 1, mm = m + l >> 1;
        if (check(m) >= check(mm))r = m;
        else l = mm;
    }
    ull ans = 2e18;
    for (ll i = l; i <= r; i++) ans = min(ans, check(i));
    write(ans);
    return 0;
}