/**
* A really nice dp + greedy problem~ I can recall that this is one the first few IOI
* problems I ever saw, and I also heard the solution from tinjyu.
*
* Anyway, the first few subtasks are pretty easy. In particular, when there's only
* one person (e.g. when y=0) we can simply sort all flavours according to price (for
* that person) and greedily pick from the small ones. From the constraints, one may
* guess that the only way to AC this problem is to do a O(nx) dp (not a O(nxy) one).
* Now, let dp[i][t] be the dp state when we only consider the first i flavours and
* exactly t dollars are spent on store B. Each dp state already captures the cost in
* B, so it may (?!) natural to maximize the num of unique flavours she can buy while
* minimizing the amount of money spent on store A.
* So, we first SORT the flavours based on their price in store A (from small to
* large). We know that if we've selected a subset of jelly in B, she would just
* greedily pick the remaining ones in store A (from left to right). Therefore, in
* each dp state, we store the max num of flavours we can buy. In addition, among all
* choices that can reach the max. num of flavours, we store the max amount of money
* that can be left with in store A. It's not difficult to prove that this is indeed
* a valid approach using greedy arguments. Nice problem!
*
* Impl2 is inspired by https://codeforces.com/blog/entry/91804 and
* https://codeforces.com/blog/entry/82625. Basically, we split the array into two
* sections where the last element in the 1st section is the last item we buy in
* store A. We use dp to compute the smallest cost used in the 1st part, and use dp
* again to compute the num of flavours in 2nd part. The idea is to split the items
* into two parts AND also split our money into two.
*
* Time Complexity: O(ny) (or equivalently, O(nx))
* Implementation 2
*/
#include <bits/stdc++.h>
#include "jelly.h"
typedef std::vector<int> vec;
const int INF = 0x3f3f3f3f;
struct jelly_t {
int a, b;
};
int find_maximum_unique(int x, int y, vec a, vec b) {
int n = a.size();
std::vector<jelly_t> values(n);
for (int i = 0; i < n; i++)
values[i].a = a[i], values[i].b = b[i];
std::sort(values.begin(), values.end(),
[](const jelly_t& j1, const jelly_t& j2) {
return j1.b < j2.b;
});
// dp1[i][t] = using exactly t dollars in store A, the min amount of money needed
// to be spent on store B if we're to buy all the first i products.
std::vector<vec> dp1(n + 1, vec(x + 1));
dp1[0][0] = 0;
for (int t = 1; t <= x; t++)
dp1[0][t] = INF;
for (int k = 0; k < n; k++) {
for (int t = 0; t <= x; t++) {
dp1[k + 1][t] = dp1[k][t] + values[k].b;
if (t >= values[k].a)
dp1[k + 1][t] = std::min(dp1[k + 1][t], dp1[k][t - values[k].a]);
}
}
// dp2[i][t] = max num of flavours we can buy, with t dollars in store A.
std::vector<vec> dp2(n + 2, vec(x + 1));
dp2[n + 1][0] = 0;
for (int t = 1; t <= x; t++)
dp2[n + 1][t] = -INF;
for (int k = n; k >= 1; k--) {
for (int t = 0; t <= x; t++) {
dp2[k][t] = dp2[k + 1][t];
if (t >= values[k - 1].a)
dp2[k][t] = std::max(dp2[k][t], dp2[k + 1][t - values[k - 1].a] + 1);
}
}
// exactly -> at most
for (int k = 0; k <= n; k++) {
for (int t = 1; t <= x; t++)
dp1[k][t] = std::min(dp1[k][t], dp1[k][t - 1]);
}
int ans = 0;
for (int k = 0; k <= n; k++) {
for (int t = 0; t <= x; t++) {
if (dp1[k][t] <= y)
ans = std::max(ans, k + dp2[k + 1][x - t]);
}
}
return ans;
}
