AtCoder Regular Contest 125 E Snack

// Problem: E - Snack
// Contest: AtCoder - AtCoder Regular Contest 125
// URL: https://atcoder.jp/contests/arc125/tasks/arc125_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;

ll n, m, a[maxn], c[maxn], d[maxn], e[maxn];
struct node {
	ll a, b;
} b[maxn];

inline bool operator < (const node &a, const node &b) {
	return a.b * b.a < b.b * a.a;
}

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
	}
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &b[i].a);
	}
	for (int i = 1; i <= m; ++i) {
		scanf("%lld", &b[i].b);
	}
	sort(a + 1, a + n + 1);
	sort(b + 1, b + m + 1);
	for (int i = 1; i <= n; ++i) {
		c[i] = c[i - 1] + a[i];
	}
	for (int i = 1; i <= m; ++i) {
		d[i] = d[i - 1] + b[i].a;
		e[i] = e[i - 1] + b[i].b;
	}
	ll ans = 1e18;
	for (int i = 0; i <= n; ++i) {
		int l = 1, r = m, p = 0;
		while (l <= r) {
			int mid = (l + r) >> 1;
			if (b[mid].b <= i * b[mid].a) {
				p = mid;
				l = mid + 1;
			} else {
				r = mid - 1;
			}
		}
		// printf("%d %d %lld\n", i, p, c[n - i] + e[p] + (d[m] - d[p]) * i);
		ans = min(ans, c[n - i] + e[p] + (d[m] - d[p]) * i);
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}

/*
很棒的 flow 题，考虑建二分图
源点向每种零食连边权为 a_i 的边，每种零食向每个孩子连边权为 b_j 的边，每个孩子向汇点连边权为 c_j 的边
这个图的最大流就是答案
直接跑最大流肯定 T，考虑最大流等价于求这个图的最小割
转而求最小割
发现如果钦定每个零食归 S 还是归 T，那么每个孩子要么断掉所有连向它的归 S 的边，要么断掉它连 T 的边
设归 S 的零食个数为 x，每个孩子的最小代价即 min(c_j, b_j * x)
既然代价只跟 x 有关了，那我们就枚举 x，零食肯定把最小的 n - x 个割掉；
把所有孩子按照 c_j/b_j 排序，c_j >= b_j * x 的孩子代价就为 b_j * x，否则为 c_j
做完了！太妙了
*/